Question

In a college classroom some chairs are designed for lef-handers. Suppose this classroom has 20 chairs, and 8 of them are made for lefies. If 5 students randomly select chairs and sit on them in this classroom, what is the probability that at least 4 of those selected will be seats for lefies?

Answer 1

Answer: 0.0578

Step-by-step explanation:

Given : Number of chairs = 20

Number of chairs made for lefties =8

Number of chairs made for righties= 20-8=12

Now, total number of way for selecting 5 chairs from 20  =$^{20}C_5=\dfrac{20!}{5!(20-5)!}$

$=\dfrac{20\times19\times18\times17\times16\times15!}{(120)15!}=15504$

$[\because\ ^nC_r=\dfrac{n!}{r!(n-r)!}]$

Number of ways to select at least 4 seats for lefies :-

$^8C_4\times^{12}C_1+^8C_5\times ^{12}C_0\\\\=\dfrac{8!}{4!4!}\cdot12+\dfrac{8!}{5!3!}(1)=896$

Now, the probability that at least 4 of those selected will be seats for lefties :-

$\dfrac{896}{15504}=0.0577915376677\approx0.0578$

hence, the required probability = 0.0578