A random variable X has the following probability distribution: X 0 1 2 3 4 P X k 2 k 4 k 6 k 8 k The value of P 1 x 4 x ≤ 2 is equal to

Question

A random variable X has the following probability distribution: X 0 1 2 3 4 P X k 2 k 4 k 6 k 8 k The value of P 1 x 4 x ≤ 2 is equal to, 4 7, 2 3, 3 7, 4 5

MARKS App · Accepted Answer

To find P 1 x 4 x ≤ 2 or P A B We know that P A B = P A ∩ B P B Given x 0 1 2 3 4 P x k 2 k 4 k 6 k 8 k P A = 2 , 3 P B = 0 , 1 , 2 P A ∩ B = P x = 2 P B = P x = 0 + P x = 1 + P x = 2 So P A B = P x = 2 P x = 0 + P x = 1 + P x = 2 = 4 k k + 2 k + 4 k = 4 k 7 k = 4 7 Hence option A is correct.

$X$	$0$	$1$	$2$	$3$	$4$
$P (X)$	$k$	$2 k$	$4 k$	$6 k$	$8 k$

$x$	$0$	$1$	$2$	$3$	$4$
$P (x)$	$k$	$2 k$	$4 k$	$6 k$	$8 k$

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