Q.A die is thrown 6 times. If 'getting an odd number' is a success, find the probability of

A die is thrown $n=6$ times and success is 'getting an odd number', so $p=\dfrac{3}{6}=\dfrac12$, $q=\dfrac12$, and $X\sim B\left(6,\tfrac12\right)$ with $P(X=x)={}^{6}C_x\left(\tfrac12\right)^6$.

1. Exactly 5 successes: $P(X=5)={}^{6}C_5\left(\tfrac12\right)^6=6\times\dfrac1{64}=\dfrac{6}{64}=\dfrac{3}{32}$.
2. At least 5 successes: $P(X\ge5)=P(5)+P(6)={}^{6}C_5\left(\tfrac12\right)^6+{}^{6}C_6\left(\tfrac12\right)^6=(6+1)\times\dfrac1{64}=\dfrac{7}{64}$.
3. At most 5 successes: this is the complement of 'all 6 odd', so $P(X\le5)=1-P(X=6)=1-{}^{6}C_6\left(\tfrac12\right)^6=1-\dfrac1{64}=\dfrac{63}{64}$.
   ✓Final answer

   $P(X=5)=\dfrac{3}{32}$, $P(X\ge5)=\dfrac{7}{64}$, $P(X\le5)=\dfrac{63}{64}$.

Identify success = odd number so p = 1/2 on a fair die, form X~B(6,1/2), then read 'exactly' as a single p.m.f. term, 'at least' as a sum of the top terms, and 'at most' as 1 minus the complementary tail.

Forgetting that all six faces make p exactly 1/2 for 'odd number' (3 odd faces out of 6); and computing 'at most 5' by summing five separate terms instead of using the much shorter complement 1 - P(X=6).