Kent's Tents has 5 green knapsacks and 3 yellow ones in stock. Curt selects 3 of them at random. Let X be the number of green knapsacks he selects. Find $P ( X \leq 2 )$ .
A) $P ( X \leq 2 ) = 0.018$
B) $P ( X \leq 2 ) = 0.821$
C) $P ( X \leq 2 ) = 0.536$
D) $P ( X \leq 2 ) = 0.179$
E) $P ( X \leq 2 ) = 0.268$

Question

Kent's Tents has 5 green knapsacks and 3 yellow ones in stock. Curt selects 3 of them at random. Let  X be the number of green knapsacks he selects. Find $P ( X \leq 2 )$ .  
A)  $P ( X \leq 2 ) = 0.018$
B)   $P ( X \leq 2 ) = 0.821$
C)   $P ( X \leq 2 ) = 0.536$
D)   $P ( X \leq 2 ) = 0.179$
E)   $P ( X \leq 2 ) = 0.268$

Quizplus · Accepted Answer

To find $P(X \leq 2)$, we calculate the probability of Curt selecting 0, 1, or 2 green knapsacks out of the 3 he chooses. The total number of ways to choose 3 knapsacks from 8 (5 green + 3 yellow) is $\binom{8}{3} = 56$. The ways to choose 0, 1, or 2 green knapsacks are as follows:- 0 green (all yellow): $\binom{5}{0} \times \binom{3}{3} = 1$ way.- 1 green: $\binom{5}{1} \times \binom{3}{2} = 5 \times 3 = 15$ ways.- 2 green: $\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30$ ways.Adding these up gives 1 + 15 + 30 = 46 ways to choose 0, 1, or 2 green knapsacks. Thus, $P(X \leq 2) = \frac{46}{56} = 0.821$.

Kent's Tents Has 5 Green Knapsacks and 3 Yellow Ones P(X≤2)P ( X \leq 2 )P(X≤2)

Kent's Tents Has 5 Green Knapsacks and 3 Yellow Ones $P ( X \leq 2 )$