Question

Determine the probability of getting the sum 6 exactly 3 times in 7 throws with a pair of fair dice

Answer 1

The sample space for throwing a pair of fair dice

= {(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}

Let A = The sum of the result when rolling two dice is 6.

Total number of possible outcomes = 36

Number of possible outcome of getting a sum of 6 when rolling two dice = 5

$P(A) = \frac{5}{36}=0.1389$

This is a binomial distribution because:

(1) The experiment is performed a fixed number of times (n=7).

(2) Each trial results in only two possible outcomes, labeled as success (The sum of the two dice is 6.) and failure (The sum of the two dice is not 6.).

(3) The trials are independent.

(4) The probability of a success (p=0.1389) in each trial remains constant.

Number of trials n=7

Probability of success p=0.1389

Let x be the number of successes.

$P(X=3) = \binom{7}{3} (0.1389)^3(1-0.1389)^{7-3} \\ = \frac{7!}{3!(7-3)!} \times (0.1389)^3(0.8611)^4 \\ = 0.0515$

The probability of getting the sum 6 exactly 3 times in 7 throws with a pair of fair dice is 0.0515.