JEE Advanced · Mathematics · 32. Probability

Three numbers are chosen at random, one after another with replacement, from the set $S=\{1,2,3, \ldots, 100\}$. Let $p_{1}$ be the probability that the maximum of chosen numbers is at least 81 and $p_{2}$ be the probability that the minimum of chosen numbers is at most 40 .
The value of $\frac{625}{4} p_{1}$ is

A 76.25
B 78.12
C 15.55
D 29

Verified Solution

Answer & Solution

Correct Answer

(A) 76.25

Step-by-step Solution

Detailed explanation

Maximum of the chosen numbers is at least

81

It means we have to choose at least one number from

81

100

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let

α, β

and

γ

be real numbers. Consider the following system of linear equations

x + 2 y + z = 7

x + α z = 11

2 x - 3 y + β z = γ

Match each entry in List-I to the correct entries in List-II.

	List-I		List-II
$(P)$	If $β = \frac{1}{2} (7 α - 3)$ and $γ = 28$ then the system has	$(1)$	a unique solution
$(Q)$	If $β = \frac{1}{2} (7 α - 3)$ and $γ \neq 28$ , then the system has	$(2)$	no solution
$(R)$	If $β \neq \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ \neq 28$ , then the system has	$(3)$	infinitely many solutions
$(S)$	If $β \neq \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ = 28$ , then the system has	$(4)$	$x = 11, y = - 2$ and $z = 0$ as a solution
		$(5)$	$x = - 15, y = 4$ and $z = 0$ as a solution

The correct option is:

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