Question

In a random experiment, two dice are thrown and the sum of the numbers appeared on them is recorded. This experiment is repeated 9 times. If the probability that a sum of 6 appears at least once is \( P_1 \) and a sum of 8 appears at least once is \( P_2 \), then \( P_1 : P_2 = \):

• A. \( 4:3 \)
• B. \( 3:1 \)
• C. \( 1:2 \)
• D. \( 1:1 \)

Hint:

When dealing with independent events over multiple trials, use the complement rule to find the probability of an event occurring at least once, and remember that the ratio of such probabilities will be based on their individual probabilities.

Answer 1

The Correct Option is D

In this problem, we are throwing two dice and recording the sum of the numbers that appear on them. The experiment is repeated 9 times. We are given that the probability of a sum of 6 appearing at least once is \( P_1 \), and the probability of a sum of 8 appearing at least once is \( P_2 \). We are asked to find the ratio \( P_1 : P_2 \). ### Step 1: Calculate the probability of a sum of 6. The possible outcomes when two dice are thrown are 36, and the combinations that give a sum of 6 are: \[ (1, 5), (2, 4), (3, 3), (4, 2), (5, 1). \] So, the probability of getting a sum of 6 in one throw is: \[ P(\text{sum of 6}) = \frac{5}{36}. \] Thus, the probability that a sum of 6 does not appear in one throw is: \[ P(\text{not sum of 6}) = 1 - \frac{5}{36} = \frac{31}{36}. \] The probability that a sum of 6 does not appear in 9 independent throws is: \[ P(\text{not sum of 6 in 9 throws}) = \left( \frac{31}{36} \right)^9. \] Thus, the probability that a sum of 6 appears at least once in 9 throws is: \[ P_1 = 1 - \left( \frac{31}{36} \right)^9. \] ### Step 2: Calculate the probability of a sum of 8. Similarly, the combinations that give a sum of 8 are: \[ (2, 6), (3, 5), (4, 4), (5, 3), (6, 2). \] So, the probability of getting a sum of 8 in one throw is: \[ P(\text{sum of 8}) = \frac{5}{36}. \] Thus, the probability that a sum of 8 does not appear in one throw is: \[ P(\text{not sum of 8}) = 1 - \frac{5}{36} = \frac{31}{36}. \] The probability that a sum of 8 does not appear in 9 independent throws is: \[ P(\text{not sum of 8 in 9 throws}) = \left( \frac{31}{36} \right)^9. \] Thus, the probability that a sum of 8 appears at least once in 9 throws is: \[ P_2 = 1 - \left( \frac{31}{36} \right)^9. \] ### Step 3: Compare the probabilities \( P_1 \) and \( P_2 \). We observe that the probabilities \( P_1 \) and \( P_2 \) are based on the same value, \( \left( \frac{31}{36} \right)^9 \), and hence they are equal. Therefore, the ratio of \( P_1 \) to \( P_2 \) is: \[ P_1 : P_2 = 1:1. \] Thus, the correct answer is: \[ \boxed{1:1}. \]

Answer 2

Given that two dice are thrown, the sum of numbers appearing can be any integer from 2 to 12. Let's determine the probability for specific sums.

The pairs that give a sum of 6 are: (1,5), (2,4), (3,3), (4,2), (5,1). Thus, there are 5 successful outcomes. The sample space when throwing two dice is 6×6=36. Therefore, the probability \( P(S=6) \) is \( \frac{5}{36} \).

The pairs that give a sum of 8 are: (2,6), (3,5), (4,4), (5,3), (6,2). Thus, there are also 5 successful outcomes. Therefore, the probability \( P(S=8) \) is also \( \frac{5}{36} \).

The complement probability is used to calculate the probability of the event occurring at least once: If \( A \) is the event that a sum appears at least once in 9 trials, then the complement event \( A' \) is that the sum does not appear in all 9 trials.

Let \( \bar{P_1} \) be the probability that a sum of 6 does not appear in a single trial, so \( \bar{P_1} = 1 - \frac{5}{36} = \frac{31}{36} \). Thus, the probability that 6 does not appear in all 9 trials is \( \left(\frac{31}{36}\right)^9 \). Therefore, the probability that 6 appears at least once, \( P_1 \), is:

\( P_1 = 1 - \left(\frac{31}{36}\right)^9 \).

Similarly, the probability that a sum of 8 does not appear in a single trial is the same, \( \bar{P_2} = \frac{31}{36} \). The probability that 8 does not appear in all 9 trials is \( \left(\frac{31}{36}\right)^9 \). Thus, the probability that 8 appears at least once, \( P_2 \), is:

\( P_2 = 1 - \left(\frac{31}{36}\right)^9 \).

Since \( P_1 \) and \( P_2 \) can be expressed similarly, \( P_1 : P_2 = 1 : 1 \).