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Five dice are thrown simultaneously. If the occurrence of an even number in a single dice is considered a success, then the probability of at most 3 successes is

Solution

✅ Correct Option: 2

Five dice are thrown simultaneously. A success is getting an even number (2, 4, or 6) on a single die.

For a single die:

Probability of success (even number) =36=12= \frac{3}{6} = \frac{1}{2}

Probability of failure (odd number) =36=12= \frac{3}{6} = \frac{1}{2}


At most 3 successes means 0, 1, 2, or 3 successes.

Using the complement approach:

P(at most 3 successes) =1−= 1 - P(more than 3 successes)

=1−[P(4 successes)+P(5 successes)]= 1 - [P(4 \text{ successes}) + P(5 \text{ successes})]


Using the binomial probability formula with n=5n = 5, p=12p = \frac{1}{2}, q=12q = \frac{1}{2}:

P(4 successes) =(54)×(12)4×(12)1= \binom{5}{4} \times \left(\frac{1}{2}\right)^4 \times \left(\frac{1}{2}\right)^1

=5×116×12= 5 \times \frac{1}{16} \times \frac{1}{2}

=532= \frac{5}{32}


P(5 successes) =(55)×(12)5×(12)0= \binom{5}{5} \times \left(\frac{1}{2}\right)^5 \times \left(\frac{1}{2}\right)^0

=1×132×1= 1 \times \frac{1}{32} \times 1

=132= \frac{1}{32}


P(at most 3 successes) =1−(532+132)= 1 - \left(\frac{5}{32} + \frac{1}{32}\right)

=1−632= 1 - \frac{6}{32}

=1−316= 1 - \frac{3}{16}

=1316= \frac{13}{16}

Therefore, the probability of at most 3 successes is 1316\frac{13}{16}.

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