Question

A jar contains 6 chocolate chip cookies and 9 peanut butter cookies. Richard grabs 3 cookies at random to pack in his lunch. What is the probability that he drew 2 chocolate chip cookies and 1 peanut butter cookie?


A. $\frac{1}{54}$

B. $\frac{1}{27} }$

C. $\frac{9}{91} }$

D. $\frac{27}{91} }$

Answer 1

Here, Your Answer would be 9/91

In short, Answer is Option C

Hope this helps!

Answer 2

Answer: D. $\dfrac{27}{91}$

Step-by-step explanation:

Given: Number of chocolate chip cookies - 6

Number of peanut butter cookie = 9

Total cookies = 6+9=15

The total number of combinations of selecting 3 cookies from 15 cookies

$^{15}C_3=\dfrac{15!}{(15-3)!3!}=455$

The number of combinations to select 2 chocolate chip cookie=

$^{6}C_2\dfrac{6!}{(6-2)!2!}=15$

The number of combinations to select 1 peanut butter cookie=

$^{9}C_1=\dfrac{9!}{(9-1)!1!}=9$

The probability that he drew 2 chocolate chip cookies and 1 peanut butter cookie=$\dfrac{^{9}C_1\times ^{9}C_1}{^{15}C_3}=\dfrac{15\times9}{455}=\dfrac{27}{91}$