Bag $B_1$ contains 6 white and 4 blue balls, Bag $B_2$ contains 4 white and 6 blue balls, and Bag $B_3$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability that the ball is drawn from Bag $B_2$ is:
(A)
$\frac{2}{5}$
(B)
$\frac{4}{15}$
(C)
$\frac{1}{3}$
(D)
$\frac{2}{3}$
EASY
Correct Answer:B
Explanation
$E_1$: Bag $B_1$ is selected
$B_1$: 6 W 4 B
$B_2$: 4 W 6 B
$B_3$: 5 W 5 B
$E_2$: bag $B_2$ is selected
We have to find $P(\frac{E_2}{A})$
$P(\frac{E_2}{A}) = \frac{P(E_2)P(\frac{A}{E_2})}{P(E_1)P(\frac{A}{E_1}) + P(E_2)P(\frac{A}{E_2}) + P(E_3)P(\frac{A}{E_3})} = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}} = \frac{4}{15}$
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MCQ_SINGLE
4 Mark(s)
2025
JEE Main 2025 (Online) 28th January Evening Shift
Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is:
(A)
$\frac{1}{4}$
(B)
$\frac{1}{2}$
(C)
$\frac{1}{3}$
(D)
$\frac{2}{3}$
EASY
Correct Answer:B
Explanation
A, E,G R D N
Probabllity $(P) = \frac{\text{favourable case}}{\text{Total case}}$
(when A & E are in order)
Total case = $6!$
Favourable case = ${6}C_2 . 4! = (15)4! = (30)4!$
$P = \frac{(15)4!}{(30)4!} = \frac{1}{2}$
Probability when not in order = $1 - \frac{1}{2} = \frac{1}{2}$