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 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}