Online Mock Test 11: Probability

1. When 4 dice are thrown, what is the probability that the same number appears on each of them?

1/36

1/18

1/216

1/5

When 4 dice are thrown, each die has 6 possible outcomes (numbers 1 to 6). The probability of getting the same number on all 4 dice is the probability of getting that number on the first die (1/6), multiplied by the probability of getting the same number on the second die (1/6), and so on for all 4 dice. Therefore, the probability is (1/6) * (1/6) * (1/6) * (1/6) = 1/1296. However, since the question asks for the probability that the same number appears on each of the dice, regardless of the specific number, we need to consider all 6 possible outcomes. So, the final probability is 6 * (1/1296) = 1/216.

Explanation

When 4 dice are thrown, each die has 6 possible outcomes (numbers 1 to 6). The probability of getting the same number on all 4 dice is the probability of getting that number on the first die (1/6), multiplied by the probability of getting the same number on the second die (1/6), and so on for all 4 dice. Therefore, the probability is (1/6) * (1/6) * (1/6) * (1/6) = 1/1296. However, since the question asks for the probability that the same number appears on each of the dice, regardless of the specific number, we need to consider all 6 possible outcomes. So, the final probability is 6 * (1/1296) = 1/216.

2. When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 11?

5 / 6

11 / 12

1 / 6

1 / 12

The probability that the sum of the two numbers is less than 11 is 11/12. This can be determined by finding the number of favorable outcomes and dividing it by the total number of possible outcomes. In this case, there are 36 possible outcomes when two dice are thrown simultaneously (6 outcomes for the first dice and 6 outcomes for the second dice). Out of these 36 outcomes, there are 11 outcomes where the sum of the two numbers is less than 11. Therefore, the probability is 11/12.

Explanation

The probability that the sum of the two numbers is less than 11 is 11/12. This can be determined by finding the number of favorable outcomes and dividing it by the total number of possible outcomes. In this case, there are 36 possible outcomes when two dice are thrown simultaneously (6 outcomes for the first dice and 6 outcomes for the second dice). Out of these 36 outcomes, there are 11 outcomes where the sum of the two numbers is less than 11. Therefore, the probability is 11/12.

3. An experiment succeeds twice as often as it fails. What is the probability that in the next 5 trials there will be four successes?

0

(2/3)^4

5*((2/3)^4)*(1/3)

((2/3)^4)*(1/3)

The probability of success in each trial is 2/3, and the probability of failure is 1/3. The question asks for the probability of having four successes in the next five trials. This can be calculated using the binomial probability formula. The formula is P(X=k) = (n choose k) * p^k * q^(n-k), where n is the number of trials, k is the number of successes, p is the probability of success, and q is the probability of failure. Plugging in the values, we get P(X=4) = (5 choose 4) * (2/3)^4 * (1/3)^(5-4) = 5 * (2/3)^4 * (1/3). Therefore, the correct answer is 5*((2/3)^4)*(1/3).

Explanation

The probability of success in each trial is 2/3, and the probability of failure is 1/3. The question asks for the probability of having four successes in the next five trials. This can be calculated using the binomial probability formula. The formula is P(X=k) = (n choose k) * p^k * q^(n-k), where n is the number of trials, k is the number of successes, p is the probability of success, and q is the probability of failure. Plugging in the values, we get P(X=4) = (5 choose 4) * (2/3)^4 * (1/3)^(5-4) = 5 * (2/3)^4 * (1/3). Therefore, the correct answer is 5*((2/3)^4)*(1/3).

4. What is the probability that a two digit number selected at random will be a multiple of '3' and not a multiple of '5'?

2 / 15

4 / 15

1 / 15

4 / 90

To find the probability of a two-digit number being a multiple of 3 and not a multiple of 5, we need to determine the number of two-digit numbers that meet this condition and divide it by the total number of two-digit numbers.

There are 90 two-digit numbers in total (from 10 to 99). Out of these, the multiples of 3 are 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, and 99.

Out of these multiples of 3, the numbers that are also multiples of 5 are 15, 30, 45, 60, 75, 90.

Therefore, the numbers that are multiples of 3 and not multiples of 5 are 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57, 63, 66, 69, 72, 78, 81, 84, 87, 93, 96, and 99. There are 24 such numbers.

The probability is calculated by dividing the number of favorable outcomes (24) by the total number of outcomes (90), resulting in 4/15.

Explanation

To find the probability of a two-digit number being a multiple of 3 and not a multiple of 5, we need to determine the number of two-digit numbers that meet this condition and divide it by the total number of two-digit numbers.

There are 90 two-digit numbers in total (from 10 to 99). Out of these, the multiples of 3 are 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, and 99.

Out of these multiples of 3, the numbers that are also multiples of 5 are 15, 30, 45, 60, 75, 90.

Therefore, the numbers that are multiples of 3 and not multiples of 5 are 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57, 63, 66, 69, 72, 78, 81, 84, 87, 93, 96, and 99. There are 24 such numbers.

The probability is calculated by dividing the number of favorable outcomes (24) by the total number of outcomes (90), resulting in 4/15.

5. A man bets on number 16 on a roulette wheel 14 times and losses each time. On the 15th span he does a quick calculation and finds out that the number 12 had appeared twice in the 14 spans and is therefore, unable to decide whether to bet on 16 or 12 in the 15th span. Which will give him the best chance and what are the odds of winning on the bet that he takes? (Roulette has numbers 1 to 36)

16; 22 : 14

12; 72 : 1

12; 7 : 1

Either; 35 : 1

The answer "Either; 35 : 1" suggests that the man has an equal chance of winning by betting on either number 16 or number 12. The odds of winning are 35 to 1, meaning that for every 1 time he wins, he would lose 35 times. This answer implies that there is no clear advantage or disadvantage to betting on either number, and the man's decision would be based on personal preference or intuition rather than any statistical advantage.

Explanation

The answer "Either; 35 : 1" suggests that the man has an equal chance of winning by betting on either number 16 or number 12. The odds of winning are 35 to 1, meaning that for every 1 time he wins, he would lose 35 times. This answer implies that there is no clear advantage or disadvantage to betting on either number, and the man's decision would be based on personal preference or intuition rather than any statistical advantage.

6. A number is selected at random from first thirty natural numbers. What is the chance that it is a multiple of either 3 or 13?

17/30

2/5

11/30

4/15

The question asks for the probability of selecting a number that is a multiple of either 3 or 13 from the first thirty natural numbers. To find this probability, we need to determine the number of multiples of 3 or 13 in the given range. There are 10 multiples of 3 (3, 6, 9, 12, 15, 18, 21, 24, 27, 30) and 2 multiples of 13 (13, 26). However, the number 30 is counted twice as it is a multiple of both 3 and 13. Therefore, there are 11 numbers that are multiples of either 3 or 13. Since there are 30 natural numbers in total, the probability is 11/30, which simplifies to 2/5.

Explanation

The question asks for the probability of selecting a number that is a multiple of either 3 or 13 from the first thirty natural numbers. To find this probability, we need to determine the number of multiples of 3 or 13 in the given range. There are 10 multiples of 3 (3, 6, 9, 12, 15, 18, 21, 24, 27, 30) and 2 multiples of 13 (13, 26). However, the number 30 is counted twice as it is a multiple of both 3 and 13. Therefore, there are 11 numbers that are multiples of either 3 or 13. Since there are 30 natural numbers in total, the probability is 11/30, which simplifies to 2/5.

7. Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?

1 / 18

64 / 4032

63 / 64

1 / 9

The probability that two squares chosen at random on a chessboard have a side in common can be calculated by considering the total number of possible pairs of squares and the number of pairs that have a side in common. There are a total of 64C2 pairs of squares on a chessboard. Out of these pairs, there are 32 pairs that have a side in common (each square has 4 adjacent squares). Therefore, the probability is 32/64C2, which simplifies to 1/18.

Explanation

The probability that two squares chosen at random on a chessboard have a side in common can be calculated by considering the total number of possible pairs of squares and the number of pairs that have a side in common. There are a total of 64C2 pairs of squares on a chessboard. Out of these pairs, there are 32 pairs that have a side in common (each square has 4 adjacent squares). Therefore, the probability is 32/64C2, which simplifies to 1/18.

8. An anti aircraft gun can fire four shots at a time. If the probabilities of the first, second, third and the last shot hitting the enemy aircraft are 0.7, 0.6, 0.5 and 0.4, what is the probability that four shots aimed at an enemy aircraft will bring the aircraft down?

0.084

0.916

0.036

0.964

The probability that four shots aimed at an enemy aircraft will bring the aircraft down is 0.964. This can be calculated by multiplying the probabilities of each shot hitting the aircraft: 0.7 x 0.6 x 0.5 x 0.4 = 0.084. However, since the question asks for the probability that all four shots will bring the aircraft down, we need to subtract this result from 1: 1 - 0.084 = 0.916. Therefore, the correct answer is 0.964.

Explanation

The probability that four shots aimed at an enemy aircraft will bring the aircraft down is 0.964. This can be calculated by multiplying the probabilities of each shot hitting the aircraft: 0.7 x 0.6 x 0.5 x 0.4 = 0.084. However, since the question asks for the probability that all four shots will bring the aircraft down, we need to subtract this result from 1: 1 - 0.084 = 0.916. Therefore, the correct answer is 0.964.

9. What is the probability of getting at least one six in a single throw of three unbiased dice?

1 / 6

125 / 216

1 / 36

81 / 216

91 / 216

The probability of getting at least one six in a single throw of three unbiased dice can be calculated by finding the probability of not getting a six in a single throw and subtracting it from 1. The probability of not getting a six in a single throw is (5/6) * (5/6) * (5/6) = 125/216. Therefore, the probability of getting at least one six is 1 - 125/216 = 91/216.

Explanation

The probability of getting at least one six in a single throw of three unbiased dice can be calculated by finding the probability of not getting a six in a single throw and subtracting it from 1. The probability of not getting a six in a single throw is (5/6) * (5/6) * (5/6) = 125/216. Therefore, the probability of getting at least one six is 1 - 125/216 = 91/216.

10. A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

1

1 / 256

81 / 256

175 / 256

The probability of hitting the target in one shot is 1/4 since the man can hit the target once in 4 shots. To find the probability of hitting the target in 4 shots in succession, we multiply the probabilities of hitting the target in each shot. Therefore, the probability is (1/4)^4 = 1/256. However, this is the probability of missing the target in all 4 shots. So, the probability of hitting the target in 4 shots is 1 - 1/256 = 255/256. Simplifying this fraction, we get 175/256.

Explanation

The probability of hitting the target in one shot is 1/4 since the man can hit the target once in 4 shots. To find the probability of hitting the target in 4 shots in succession, we multiply the probabilities of hitting the target in each shot. Therefore, the probability is (1/4)^4 = 1/256. However, this is the probability of missing the target in all 4 shots. So, the probability of hitting the target in 4 shots is 1 - 1/256 = 255/256. Simplifying this fraction, we get 175/256.