Multiple choice

A die is thrown a fixed number of times. If probability of getting even number $3$ times is same as the probability of getting even number $4$ times, then probability of getting even number exactly once is

  1. $\dfrac{1}{4}$
  2. $\dfrac{3}{128}$
  3. $\dfrac{5}{64}$
  4. $\dfrac{7}{128}$
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D Correct answer
Explanation

For a binomial distribution, P(X=3) = P(X=4). This implies nC3 * p^3 * q^(n-3) = nC4 * p^4 * q^(n-4). Simplifying gives q/n-3 = p/4. With p=1/2, n=7. The probability of getting an even number exactly once is 7C1 * (1/2)^1 * (1/2)^6 = 7/128.

AI explanation

Using the binomial distribution formula, the probability of getting an even number is p and an odd number is q. Since nC3 * p^3 * q^(n-3) = nC4 * p^4 * q^(n-4), we get p/q = (n-3)/4. Because p equals q for a die, p = 1/2, which makes n = 7. The probability of getting an even number exactly once is 7C1 * (1/2)^1 * (1/2)^6. Calculating 7/128 gives the result 7/128.

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