Question

A bag contains $6$ red balls and $7$ green balls. You plan to select $5$ balls at random. What is the probability of selecting (without replacement) $5$ green balls?

Answer 1

Hint : Here, in the given question, we are given that a bag contains $6$ red balls and $7$ green balls, and we need to find the probability of selecting $5$ green balls, without replacement. Here, ``without replacement" means that you don't put the ball or balls back in the box so that the number of balls in the box gets less as each ball is removed. We will first find the number of ways of selecting five balls from $13$ balls and after that we find the number of ways of selecting five green balls from $7$ green balls. At the end we will divide them to get the probability of selecting $5$ green balls, without replacement.

Complete step-by-step answer :
Given:
Red balls = $6$
Green balls = $7$
Therefore, total balls = $6 + 7 = 13$
Five balls can be taken from $13$ balls in ${}^{13}{C_5}$ ways.
As we know, ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$. Therefore,
$\Rightarrow {}^{13}{C_5} = \dfrac{{13!}}{{\left( {13 - 5} \right)!5!}}$
On expanding the factorial, we get
$\Rightarrow {}^{13}{C_5} = \dfrac{{13 \times 12 \times 11 \times 10 \times 9 \times 8!}}{{8! \times 5 \times 4 \times 3 \times 2 \times 1}}$
On canceling-out common terms, we get
$\Rightarrow {}^{13}{C_5} = \dfrac{{154440}}{{120}}$
On division, we get
$\Rightarrow {}^{13}{C_5} = 1287$
Five green balls can be taken from $7$ green balls in ${}^7{C_5}$ ways.
$\Rightarrow {}^7{C_5} = \dfrac{{7!}}{{\left( {7 - 5} \right)! \times 5!}}$
On expanding the factorial, we get
$\Rightarrow {}^7{C_5} = \dfrac{{7 \times 6 \times 5!}}{{2! \times 5!}}$
On canceling-out common terms, we get
$\Rightarrow {}^7{C_5} = \dfrac{{7 \times 6}}{2}$
$\Rightarrow {}^7{C_5} = 21$
Probability of taking $5$ green balls
$\dfrac{{{}^{13}{C_5}}}{{{}^7{C_5}}} = \dfrac{{21}}{{1287}}$
On division, we get
$\dfrac{{{}^{13}{C_5}}}{{{}^7{C_5}}} = 0.016$
So, the correct answer is “$0.016$.”.

Note : Students must be careful between how to find probability with and without replacement. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick. Without replacement: When sampling is done without replacement, each member of a population may be chosen only once.