Four defective oranges are accidentally mixed | vedclass

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(B) Total oranges = $4 + 16 = 20$. Three oranges are drawn. The number of ways to draw $3$ oranges from $20$ is $C(20, 3) = \frac{20 	imes 19 	imes

Vedclass · Accepted Answer

$\begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 \ \hline P(X=x) & \frac{28}{57} & \frac{8}{19} & \frac{8}{95} & \frac{1}{285} \ \hline \end{array}$

Vedclass · Answer

(B) Total oranges = $4 + 16 = 20$. Three oranges are drawn. The number of ways to draw $3$ oranges from $20$ is $C(20, 3) = \frac{20 	imes 19 	imes 18}{3 	imes 2 	imes 1} = 1140$.Let $X$ be the number of defective oranges. $X$ can take values $0, 1, 2, 3$.$P(X=0) = \frac{C(4,0) 	imes C(16,3)}{1140} = \frac{1 	imes 560}{1140} = \frac{560}{1140} = \frac{28}{57}$.$P(X=1) = \frac{C(4,1) 	imes C(16,2)}{1140} = \frac{4 	imes 120}{1140} = \frac{480}{1140} = \frac{24}{57} = \frac{8}{19}$.$P(X=2) = \frac{C(4,2) 	imes C(16,1)}{1140} = \frac{6 	imes 16}{1140} = \frac{96}{1140} = \frac{8}{95}$.$P(X=3) = \frac{C(4,3) 	imes C(16,0)}{1140} = \frac{4 	imes 1}{1140} = \frac{4}{1140} = \frac{1}{285}$.Comparing with the options,option $B$ matches these values.

$X = x_{i}$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x_{i})$	$0.1$	$k$	$0.2$	$2k$	$3k$	$k$

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is

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