Online Mock Test 11: Probability

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.

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.

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.

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.

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.

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.

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).

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.

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.

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.