A die is thrown 4 times and getting 3 is considered a success. The probability of 2 successes is:

A die is thrown 4 times, and getting 3 is considered a success. The probability of exactly 2 successes needs to be found.

Total throws (n) = 4

Successes desired (k) = 2

Probability of getting 3 in one throw = $\frac{1}{6}$

Probability of not getting 3 = $\frac{5}{6}$

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This is a binomial probability problem. The formula is:

$P(\text{exactly 2 successes}) = \binom{4}{2} \times \left(\frac{1}{6}\right)^2 \times \left(\frac{5}{6}\right)^2$

where:

• $\binom{4}{2}$ = number of ways to choose which 2 throws will be successes
• $\left(\frac{1}{6}\right)^2$ = probability of getting 3 twice
• $\left(\frac{5}{6}\right)^2$ = probability of not getting 3 in the other 2 throws

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Calculate $\binom{4}{2}$:

$\binom{4}{2} = \frac{4!}{2! \times 2!}$

$= \frac{4 \times 3}{2 \times 1}$

$= \frac{12}{2}$

$= 6$

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Substitute into the formula:

$P = 6 \times \left(\frac{1}{6}\right)^2 \times \left(\frac{5}{6}\right)^2$

$P = 6 \times \frac{1}{36} \times \frac{25}{36}$

$P = 6 \times \frac{25}{1296}$

$P = \frac{150}{1296}$

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Simplify by dividing both numerator and denominator by 6:

$P = \frac{150 \div 6}{1296 \div 6}$

$P = \frac{25}{216}$

Therefore, the probability of 2 successes is $\frac{25}{216}$.