Exercise 14.1: Basic Probability
Question 1:
1. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = .
(ii) The probability of an event that cannot happen is . Such an event is
called .
(iii) The probability of an event that is certain to happen is . Such an event
is called .
(iv) The sum of the probabilities of all the elementary events of an experiment is
.
(v) The probability of an event is greater than or equal to and less than or
equal to .
Solution :-
Here are the completed statements:
(i) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.
2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:-
(i) A driver attempts to start a car. The car starts or does not start.
- Outcomes: "Car starts" or "Car does not start."
- Explanation: These outcomes are not equally likely, as the likelihood of the car starting depends on factors like the car's condition, fuel availability, etc. If the car is in good condition, the probability of starting is higher than not starting.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
- Outcomes: "Shoots successfully" or "Misses the shot."
- Explanation: These outcomes are not equally likely, as the probability depends on the player's skill, distance, and other conditions. A skilled player is more likely to score than miss.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
- Outcomes: "Answer is right" or "Answer is wrong."
- Explanation: These outcomes are equally likely, assuming the person answering has no knowledge of the question and is guessing. In this case, the probability of each outcome is 50% (1/2).
(iv) A baby is born. It is a boy or a girl.
- Outcomes: "Baby is a boy" or "Baby is a girl."
- Explanation: These outcomes are approximately equally likely, as the probability of having a boy or a girl is roughly the same (close to 50%) in most populations. However, slight biological and statistical variations may exist in certain cases.
Final Answer:
- Equally likely outcomes: (iii), (iv)
- Not equally likely outcomes: (i), (ii)
3. Why is tossing a coin considered to be a fair way of deciding which team should get the
ball at the beginning of a football game?
Solution :-
Tossing a coin is considered a fair way to decide which team should get the ball at the beginning of a football game because:
-
Equally Likely Outcomes: A coin has two faces — heads and tails — and if the coin is fair (not biased), the probability of getting heads or tails is equal, i.e., 50% (1/2) for each outcome.
-
Randomness: The result of a coin toss is random and not influenced by any external factors like skill, strategy, or bias, ensuring impartiality.
-
Neutrality: Neither team has control over the outcome of the toss, and it provides both teams an equal chance to win the toss.
-
Universally Accepted Method: Tossing a coin is a widely accepted and simple method for making decisions where fairness is required, as it relies on chance rather than favoritism or subjective judgment.
4. Which of the following cannot be the probability of an event?
(A) 2/3 (B) –1.5 (C) 15% (D) 0.7
Solution :-
The correct answer is: (B) –1.5
Explanation:
The probability of an event must always lie within the range 0 ≤ Probability ≤ 1. This means:
- Probabilities cannot be negative.
- Probabilities cannot exceed 1.
Now let's analyze each option:
- (A) 2/3: This is valid because it lies between 0 and 1.
- (B) –1.5: This is invalid because probabilities cannot be negative.
- (C) 15%: This is valid because 15% = 0.15, which lies between 0 and 1.
- (D) 0.7: This is valid because it lies between 0 and 1.
Thus, the probability –1.5 is not valid.
5. If P(E) = 0.05, what is the probability of ‘not E’?
Solution:-
The probability of ‘not E’ can be calculated using the formula:
P(not E) = 1 - P(E)
In this case, the probability of event E is given as 0.05.
So,
P(not E) = 1 - 0.05 = 0.95
Final Answer:
The probability of ‘not E’ is 0.95.
6. A bag contains lemon flavoured candies only. Malini takes out one candy without
looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution :-
A bag contains only lemon-flavored candies. Malini takes out one candy without looking into the bag. Let’s find the probabilities:
(i) Probability of taking out an orange-flavored candy:
Since the bag contains only lemon-flavored candies, it is impossible to take out an orange-flavored candy.
Probability = 0
(ii) Probability of taking out a lemon-flavored candy:
Since all the candies in the bag are lemon-flavored, it is certain that Malini will take out a lemon-flavored candy.
Probability = 1
Final Answer:
- Probability of taking out an orange-flavored candy = 0
- Probability of taking out a lemon-flavored candy = 1
7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:-
It is given that the probability of 2 students not having the same birthday is 0.992.
To find the probability that the 2 students have the same birthday, we use the formula:
P(having the same birthday) = 1 - P(not having the same birthday)
Substitute the given value:
P(having the same birthday) = 1 - 0.992 = 0.008
Final Answer:
The probability that the 2 students have the same birthday is 0.008.
8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag.
What is the probability that the ball drawn is (i) red ? (ii) not red?
Solution:-
A bag contains 3 red balls and 5 black balls. The total number of balls in the bag is:
3 (red) + 5 (black) = 8 balls
Let’s calculate the probabilities:
(i) Probability that the ball drawn is red:
The number of red balls is 3. The probability is:
P(red) = (Number of red balls) / (Total number of balls)
P(red) = 3 / 8
(ii) Probability that the ball drawn is not red:
The number of balls that are not red (black balls) is 5. The probability is:
P(not red) = (Number of black balls) / (Total number of balls)
P(not red) = 5 / 8
Final Answer:
- Probability of drawing a red ball = 3/8
- Probability of drawing a ball that is not red = 5/8
9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken
out of the box at random. What is the probability that the marble taken out will be
(i) red ? (ii) white ? (iii) not green?
Solution:-
A box contains the following marbles:
- 5 red marbles
- 8 white marbles
- 4 green marbles
The total number of marbles is:
5 (red) + 8 (white) + 4 (green) = 17 marbles
Now, let’s calculate the probabilities:
(i) Probability that the marble taken out is red:
The number of red marbles is 5. The probability is:
P(red) = (Number of red marbles) / (Total number of marbles)
P(red) = 5 / 17
(ii) Probability that the marble taken out is white:
The number of white marbles is 8. The probability is:
P(white) = (Number of white marbles) / (Total number of marbles)
P(white) = 8 / 17
(iii) Probability that the marble taken out is not green:
The number of marbles that are not green (red + white) is 5 + 8 = 13. The probability is:
P(not green) = (Number of marbles that are not green) / (Total number of marbles)
P(not green) = 13 / 17
Final Answer:
- Probability of drawing a red marble = 5/17
- Probability of drawing a white marble = 8/17
- Probability of drawing a marble that is not green = 13/17
10. A piggy bank contains hundred 50p coins, fifty ` 1 coins, twenty ` 2 coins and ten ` 5
coins. If it is equally likely that one of the coins will fall out when the bank is turned
upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be
a ` 5 coin?
Solution:-
A piggy bank contains the following coins:
- 100 50p coins
- 50 `1 coins
- 20 `2 coins
- 10 `5 coins
The total number of coins in the piggy bank is:
100 + 50 + 20 + 10 = 180 coins
Now, let’s calculate the probabilities:
(i) Probability that the coin will be a 50p coin:
The number of 50p coins is 100. The probability is:
P(50p coin) = (Number of 50p coins) / (Total number of coins)
P(50p coin) = 100 / 180
Simplify:
P(50p coin) = 5 / 9
(ii) Probability that the coin will not be a `5 coin:
The number of 5** coins is **10**. The number of coins that are not **5 coins is 180 - 10 = 170. The probability is:
P(not 5 coin) = (Number of coins that are not 5 coins) / (Total number of coins)
P(not 5 coin) = 170 / 180** Simplify: **P(not 5 coin) = 17 / 18
Final Answer:
- Probability of drawing a 50p coin = 5/9
- Probability of drawing a coin that is not a `5 coin = 17/18
11. Gopi buys a fish from a shop for his aquarium. The
shopkeeper takes out one fish at random from a
tank containing 5 male fish and 8 female fish (see
Fig. 14.4). What is the probability that the fish taken
out is a male fish?
Solution:-
Gopi buys a fish from a shop, and the shopkeeper randomly selects one fish from a tank that contains:
- 5 male fish
- 8 female fish
The total number of fish in the tank is:
5 (male) + 8 (female) = 13 fish
Now, let’s calculate the probability:
Probability that the fish taken out is a male fish:
The number of male fish is 5. The probability is:
P(male fish) = (Number of male fish) / (Total number of fish)
P(male fish) = 5 / 13
Final Answer:
The probability that the fish taken out is a male fish is 5/13.
12. A game of chance consists of spinning an arrow
which comes to rest pointing at one of the numbers
1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 14.5 ), and these are equally
likely outcomes. What is the probability that it will
point at
(i) 8 ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Solution:-
In a game of chance, an arrow is spun, and it comes to rest pointing at one of the following numbers:
1, 2, 3, 4, 5, 6, 7, 8
These are equally likely outcomes, meaning each number has the same chance of being selected.
Now, let’s calculate the probabilities:
(i) Probability that the arrow will point at 8:
Since there are 8 equally likely numbers, the probability of landing on any specific number (like 8) is:
P(8) = 1 / 8
(ii) Probability that the arrow will point at an odd number:
The odd numbers in the set are 1, 3, 5, 7. So there are 4 odd numbers. The probability is:
P(odd number) = (Number of odd numbers) / (Total number of outcomes)
P(odd number) = 4 / 8 = 1 / 2
(iii) Probability that the arrow will point at a number greater than 2:
The numbers greater than 2 are 3, 4, 5, 6, 7, 8. There are 6 numbers greater than 2. The probability is:
P(greater than 2) = (Number of numbers greater than 2) / (Total number of outcomes)
P(greater than 2) = 6 / 8 = 3 / 4
(iv) Probability that the arrow will point at a number less than 9:
All the numbers in the set are less than 9. Therefore, the probability is:
P(less than 9) = (Number of numbers less than 9) / (Total number of outcomes)
P(less than 9) = 8 / 8 = 1
Final Answer:
- Probability of landing on 8 = 1/8
- Probability of landing on an odd number = 1/2
- Probability of landing on a number greater than 2 = 3/4
- Probability of landing on a number less than 9 = 1
13. A die is thrown once. Find the probability of getting
(i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.
Solution :-
A die is thrown once, and the possible outcomes are:
1, 2, 3, 4, 5, 6
Now, let’s calculate the probabilities:
(i) Probability of getting a prime number:
The prime numbers between 1 and 6 are: 2, 3, 5. So, there are 3 prime numbers. The probability is:
P(prime number) = (Number of prime numbers) / (Total number of outcomes)
P(prime number) = 3 / 6 = 1 / 2
(ii) Probability of getting a number lying between 2 and 6:
The numbers between 2 and 6 are: 3, 4, 5. There are 3 such numbers. The probability is:
P(between 2 and 6) = (Number of numbers between 2 and 6) / (Total number of outcomes)
P(between 2 and 6) = 3 / 6 = 1 / 2
(iii) Probability of getting an odd number:
The odd numbers between 1 and 6 are: 1, 3, 5. There are 3 odd numbers. The probability is:
P(odd number) = (Number of odd numbers) / (Total number of outcomes)
P(odd number) = 3 / 6 = 1 / 2
Final Answer:
- Probability of getting a prime number = 1/2
- Probability of getting a number between 2 and 6 = 1/2
- Probability of getting an odd number = 1/2
14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour (ii) a face card (iii) a red face card
(iv) the jack of hearts (v) a spade (vi) the queen of diamonds
Solution :-
A standard deck of 52 cards contains the following suits:
- 13 hearts
- 13 diamonds
- 13 clubs
- 13 spades
Each suit contains 3 face cards: Jack, Queen, King.
Now, let’s calculate the probabilities for each case:
(i) Probability of getting a king of red colour:
The red suits are hearts and diamonds. The kings in these suits are the king of hearts and king of diamonds. So, there are 2 red kings. The probability is:
P(king of red colour) = (Number of red kings) / (Total number of cards)
P(king of red colour) = 2 / 52 = 1 / 26
(ii) Probability of getting a face card:
Face cards are Jack, Queen, King. There are 3 face cards in each suit, and there are 4 suits. So, there are:
3 face cards/suit × 4 suits = 12 face cards. The probability is:
P(face card) = (Number of face cards) / (Total number of cards)
P(face card) = 12 / 52 = 3 / 13
(iii) Probability of getting a red face card:
Red suits are hearts and diamonds, and each has 3 face cards. So, there are:
3 face cards/suit × 2 red suits = 6 red face cards. The probability is:
P(red face card) = (Number of red face cards) / (Total number of cards)
P(red face card) = 6 / 52 = 3 / 26
(iv) Probability of getting the jack of hearts:
There is only 1 jack of hearts in the deck. The probability is:
P(jack of hearts) = (Number of jack of hearts) / (Total number of cards)
P(jack of hearts) = 1 / 52
(v) Probability of getting a spade:
There are 13 spades in the deck. The probability is:
P(spade) = (Number of spades) / (Total number of cards)
P(spade) = 13 / 52 = 1 / 4
(vi) Probability of getting the queen of diamonds:
There is only 1 queen of diamonds in the deck. The probability is:
P(queen of diamonds) = (Number of queen of diamonds) / (Total number of cards)
P(queen of diamonds) = 1 / 52
Final Answer:
- Probability of getting a king of red colour = 1/26
- Probability of getting a face card = 3/13
- Probability of getting a red face card = 3/26
- Probability of getting the jack of hearts = 1/52
- Probability of getting a spade = 1/4
- Probability of getting the queen of diamonds = 1/52
15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their
face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card
picked up is (a) an ace? (b) a queen?
Solution:-
Here’s the explanation in simple text for your blog:
In this scenario, we have 5 cards:
- 10 of diamonds
- Jack of diamonds
- Queen of diamonds
- King of diamonds
- Ace of diamonds
These cards are well-shuffled and placed face down. One card is drawn at random, and we need to calculate the probabilities for each case.
(i) Probability that the card drawn is the queen:
Since there are 5 cards in total, and only one of them is the queen, the probability is:
P(queen) = (Number of queens) / (Total number of cards)
P(queen) = 1 / 5
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is:
(a) Ace:
After the queen is drawn and put aside, we are left with 4 cards:
- 10 of diamonds
- Jack of diamonds
- King of diamonds
- Ace of diamonds
Since there is only 1 Ace among the remaining 4 cards, the probability is:
P(ace) = (Number of aces) / (Total remaining cards)
P(ace) = 1 / 4
(b) Queen:
After the queen is drawn and put aside, there are no remaining queens. So, the probability of drawing a queen is:
P(queen) = (Number of queens left) / (Total remaining cards)
P(queen) = 0 / 4 = 0
Final Answer:
- Probability of drawing the queen = 1/5
- Probability of drawing an ace (after the queen is drawn) = 1/4
- Probability of drawing a queen (after the queen is drawn) = 0
16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just
look at a pen and tell whether or not it is defective. One pen is taken out at random from
this lot. Determine the probability that the pen taken out is a good one.
Solution:-
In this scenario, we have a total of 144 pens (12 defective and 132 good). We are asked to find the probability that a randomly chosen pen is good.
The total number of pens is:
12 (defective) + 132 (good) = 144 pens
Probability of drawing a good pen:
The number of good pens is 132, and the total number of pens is 144. So, the probability is:
P(good pen) = (Number of good pens) / (Total number of pens)
P(good pen) = 132 / 144 = 11 / 12
Final Answer:
The probability that the pen taken out is a good one is 11/12.
17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot.
What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb
is drawn at random from the rest. What is the probability that this bulb is not
defective ?
Solution:-
Here’s the explanation in simple text for your blog:
We have a total of 20 bulbs, 4 of which are defective and the remaining 16 are good. We need to calculate the probabilities for two scenarios:
(i) Probability that the bulb drawn is defective:
The total number of bulbs is 20, and the number of defective bulbs is 4. So, the probability of drawing a defective bulb is:
P(defective) = (Number of defective bulbs) / (Total number of bulbs)
P(defective) = 4 / 20 = 1 / 5
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now, one bulb is drawn at random from the remaining 19 bulbs. What is the probability that this bulb is not defective?
Since the first bulb drawn was not defective, we now have:
- 19 bulbs remaining
- 16 good bulbs left (because one good bulb was taken out, reducing the total of good bulbs)
So, the probability of drawing a non-defective (good) bulb from the remaining 19 bulbs is:
P(not defective) = (Number of good bulbs left) / (Total number of remaining bulbs)
P(not defective) = 16 / 19
Final Answers:
- The probability that the first bulb drawn is defective = 1/5
- The probability that the second bulb drawn (after the first non-defective bulb is not replaced) is not defective = 16/19
18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random
from the box, find the probability that it bears (i) a two-digit number (ii) a perfect
square number (iii) a number divisible by 5.
Solution:-
We have a box containing 90 discs numbered from 1 to 90. We are asked to calculate the probability for three scenarios:
(i) Probability that the drawn disc bears a two-digit number:
The two-digit numbers are from 10 to 90. So, the two-digit numbers are:
10, 11, 12, ..., 90. The total number of two-digit numbers is:
90 - 10 + 1 = 81
The probability is:
P(two-digit number) = (Number of two-digit numbers) / (Total number of discs)
P(two-digit number) = 81 / 90 = 9 / 10
(ii) Probability that the drawn disc bears a perfect square number:
Perfect square numbers between 1 and 90 are:
1, 4, 9, 16, 25, 36, 49, 64, 81. So, there are 9 perfect square numbers. The probability is:
P(perfect square number) = (Number of perfect square numbers) / (Total number of discs)
P(perfect square number) = 9 / 90 = 1 / 10
(iii) Probability that the drawn disc bears a number divisible by 5:
Numbers divisible by 5 between 1 and 90 are:
5, 10, 15, 20, ..., 90. These numbers form an arithmetic sequence with the first term 5 and the common difference 5. The number of terms in this sequence is:
90 / 5 = 18
The probability is:
P(divisible by 5) = (Number of numbers divisible by 5) / (Total number of discs)
P(divisible by 5) = 18 / 90 = 1 / 5
Final Answers:
- The probability that the drawn disc bears a two-digit number = 9/10
- The probability that the drawn disc bears a perfect square number = 1/10
- The probability that the drawn disc bears a number divisible by 5 = 1/5
19. A child has a die whose six faces show the letters as given below: draw 6 boxes letter A,B,C,D,E The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:-
In this scenario, we have a die with six faces, and each face shows a letter. The faces are labeled as:
A, B, C, D, E, and another letter (possibly F, if not specified).
We are asked to calculate the probability of getting certain letters when the die is thrown once.
(i) Probability of getting A:
There are 6 possible outcomes (one for each face of the die), and only one face shows A. So, the probability is:
P(A) = (Number of faces showing A) / (Total number of faces)
P(A) = 1 / 6
(ii) Probability of getting D:
Similarly, there is 1 face showing D, so the probability is:
P(D) = (Number of faces showing D) / (Total number of faces)
P(D) = 1 / 6
Final Answers:
- The probability of getting A = 1/6
- The probability of getting D = 1/6
20*. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is
the probability that it will land inside the circle with diameter 1m?
Solution :-
In this scenario, a die is dropped randomly onto a rectangular region, and we are asked to find the probability that it will land inside a circle with a diameter of 1 meter. The dimensions of the rectangle are provided as length = 3 meters and breadth = 2 meters.
Step 1: Calculate the Areas
-
Area of the circle: The formula for the area of a circle is:
Area = πr²,
where r is the radius. Since the diameter is 1 meter, the radius is 0.5 meters. So, the area of the circle is:
Area of the circle = π(0.5)² = π(0.25) ≈ 0.785 square meters.
-
Area of the rectangle: The area of the rectangle is calculated using the formula:
Area of the rectangle = length × breadth
Area of the rectangle = 3m × 2m = 6 square meters.
Step 2: Calculate the Probability
The probability of the die landing inside the circle is the ratio of the area of the circle to the area of the rectangle:
P(inside circle) = (Area of the circle) / (Area of the rectangle)
P(inside circle) ≈ 0.785 / 6 ≈ 0.1308
Final Answer:
The probability that the die will land inside the circle is approximately 0.1308, or about 13.08%.
21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri
will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one
pen at random and gives it to her. What is the probability that
(i) She will buy it ?
(ii) She will not buy it ?
Solution :-
Certainly! Here's a simplified version without LaTeX:
In a batch of 144 ball pens, 20 are defective, and the remaining 124 are good. Nuri will buy a pen if it's good but won't buy it if it's defective.
(i) Probability that Nuri will buy the pen:
Nuri will only buy the pen if it's good. Since there are 124 good pens out of 144 total pens, the probability that Nuri will buy a pen is 124 out of 144, or approximately 0.8611.
(ii) Probability that Nuri will not buy the pen:
Nuri will not buy the pen if it's defective. Since there are 20 defective pens out of 144 total pens, the probability that Nuri will not buy a pen is 20 out of 144, or approximately 0.1389.
In simple terms, Nuri has an 86.11% chance of buying the pen and a 13.89% chance of not buying it.
23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time.
Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses
otherwise. Calculate the probability that Hanif will lose the game.
Solution:
When a coin is tossed three times, there are 8 possible outcomes:
- HHH
- HHT
- HTH
- HTT
- THH
- THT
- TTH
- TTT
Out of these, Hanif wins only if all tosses are the same, i.e., either HHH (three heads) or TTT (three tails). So, Hanif wins in 2 outcomes.
For Hanif to lose, the outcomes must not be all the same. The remaining 6 outcomes (HHT, HTH, HTT, THH, THT, TTH) represent the cases where Hanif loses.
The probability of Hanif losing is the number of losing outcomes divided by the total number of outcomes:
Probability of losing = 6 (losing outcomes) ÷ 8 (total outcomes) = 3/4.
Conclusion:
The probability that Hanif will lose the game is 3/4 or 75%.
24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the
same experiment]
Solution:
When a die is thrown twice, there are 6 possible outcomes for each throw (1, 2, 3, 4, 5, or 6). So, when a die is thrown twice, the total number of possible outcomes is:
6 (for the first throw) × 6 (for the second throw) = 36 total possible outcomes.
(i) Probability that 5 will not come up either time:
For the first throw, the probability of not getting a 5 is 5 out of 6 (since there are 5 other numbers). The same is true for the second throw. Therefore, the probability that 5 will not come up on both throws is:
5/6 × 5/6 = 25/36
(ii) Probability that 5 will come up at least once:
To find the probability that 5 will come up at least once, it's easier to first calculate the probability that 5 does not come up at all (which we already found in part (i)) and then subtract that from 1.
So, the probability that 5 will come up at least once is:
1 - 25/36 = 11/36
Conclusion:
(i) The probability that 5 will not come up either time is 25/36.
(ii) The probability that 5 will come up at least once is 11/36.
25. Which of the following arguments are correct and which are not correct? Give reasons
for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes—two
heads, two tails or one of each. Therefore, for each of these outcomes, the
probability is 1/3
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even
number. Therefore, the probability of getting an odd number is 1/2 .
Solution:
(i) Argument 1:
This argument is incorrect.
When two coins are tossed, there are 4 possible outcomes:
- Heads on both coins (HH)
- Tails on both coins (TT)
- One head and one tail, which can happen in two ways: Head on the first coin and Tail on the second (HT), or Tail on the first coin and Head on the second (TH).
So, the correct possible outcomes are: HH, TT, HT, and TH, which makes 4 possible outcomes, not 3. Therefore, the probability of each outcome is 1/4, not 1/3.
(ii) Argument 2:
This argument is correct.
When a die is thrown, the possible outcomes are:
- Odd numbers: 1, 3, 5
- Even numbers: 2, 4, 6
There are 3 odd numbers and 3 even numbers on a standard die, so there are 2 equally likely outcomes: getting an odd number or getting an even number. Therefore, the probability of getting an odd number is 3 (odd outcomes) out of 6 (total outcomes), which simplifies to 1/2.
Conclusion:
- (i) The first argument is incorrect because there are 4 possible outcomes when tossing two coins, not 3.
- (ii) The second argument is correct because there are 3 odd numbers and 3 even numbers on a die, so the probability of getting an odd number is 1/2.
