Mathematics · Probability

JEE Main 2024 — 29 January, Shift 1 — Question 4

A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

Option A:
$\frac{5}{6}$
Option B:
$\frac{1}{6}$
Option C:
$\frac{5}{11}$
Correct
Option D:
$\frac{6}{11}$

Answer: C

Step-by-step solution

Required probability $=$ $\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{3} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{5} \times \frac{1}{6}+\ldots \ldots$

$=\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Mathematics
Chapter: Probability
Topic: Independent Events

If the mean and variance of the data 65,68,58,44 , 48,45,60, alpha, beta, 60 where alpha beta are 56 and 66.2 respectively, then…

Answer: C

Step-by-step solution

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