The selection process for a 4 member commitee is from a group of 5 men and 6 women. what is the probability that at most 2 women will be chosen ???
ANS: 43/66 But can someone explain how we get this answer ?
The formula for “at least” or “at most” cases is :
{n \choose r } \cdot p^r \cdot(1-p)^{n-r}
n = number of trials
r = number of specific events you wish to obtain
p = probability that the event will occur
q = probability that the event will not occur
Source: https://mathbitsnotebook.com/Algebra2/Probability/PBBinomialProbMostLeast.html
Here, In this question we’re ask at most thus, we need to take into account the case of where women = 2, women = 1 and women = 0
Then ,
probability that 2 women will selected out 4 memeber =
\frac{{6 \choose 2} \cdot{5 \choose 2}}{11 \choose 4}
probability that 1 women will selected out 4 memeber =
\frac{{6 \choose 1} \cdot{5 \choose 3}}{11 \choose 4}
probability that 0 women will selected out 4 memeber =
\frac{{6 \choose 0} \cdot{5 \choose 4}}{11 \choose 4}
At the end, add all the three cases :
\frac{5}{11}+\frac{2}{11}+\frac{5}{330} = \frac{43}{66}