A die is thrown 6 times. If "getting an odd number" is a "success", what is the probability of atmost 5 successes?

$\frac{63}{64}$

The correct answer is Option (3) → $\frac{63}{64}$

When a die is thrown, sample space = $\{1, 2, 3, 4, 5, 6\}$.

It has six equally likely outcomes.

Probability of success = P(getting an odd number) = $p =\frac{3}{6}=\frac{1}{2}$, so $q = 1-\frac{1}{2}=\frac{1}{2}$

As the die is thrown 6 times, so there are 6 trials.

Thus, we have a binomial distribution with $n = 6, p = \frac{1}{2}, q = \frac{1}{2}$.

$P(r) = {^nC}_rp^rq^{n-r}= {^6C}_r(\frac{1}{2})^r(\frac{1}{2})^{6-r} = {^6C}_r(\frac{1}{2})^6$ for $r=0,1,2,....,6$

Probability of getting atmost 5 successes

$= P(r ≤ 5) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5)$

$= 1 − P(6) = 1-{^6C}_6(\frac{1}{2})^6 = 1-1×\frac{1}{2^6}=1-\frac{1}{64}=\frac{63}{64}$