2024 Slot 1 - Three defective bulbs are mixed with 8 good ones. If three bulbs are drawn one by one with replacement, the probabilities of getting exactly 1 defective, more than 2 defective, no defective and more than 1 defective respectively are: | PYQs + Solutions

2024 Slot 1

Algebra

Hard

$\frac{27}{1331}, \frac{576}{1331}, \frac{243}{1331}$ and $\frac{512}{1331}$

$\frac{27}{1331}, \frac{243}{1331}, \frac{576}{1331}$ and $\frac{512}{1331}$

$\frac{576}{1331}, \frac{27}{1331}, \frac{512}{1331}$ and $\frac{243}{1331}$

$\frac{243}{1331}, \frac{576}{1331}, \frac{512}{1331}$ and $\frac{27}{1331}$

✅ Correct Option: 3

Related questions:

A die is rolled thrice. What is the probability of getting a number greater than $4$ in the first and the second throw of dice and a number less than $4$ in the third throw ?

Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be :

If the random variable $X$ has the following distribution:

$ X $	0	1	2	otherwise
$P(X)$	$k$	$2k$	$3k$	$0$

Match List-I with List-II:

List-I	List-II
(A) $k$	(I) $\frac{5}{6}$
(B) $P(X < 2)$	(II) $\frac{4}{3}$
(C) $E(X)$	(III) $\frac{1}{2}$
(D) $P(1 \leq X \leq 2)$	(IV) $\frac{1}{6}$

2024 Slot 1 PYQs

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