Gujarati
14.Probability
easy

A bag contains $4$ white and $3$ red balls. Two draws of one ball each are made without replacement. Then the probability that both the balls are red is

A

$\frac{1}{7}$

B

$\frac{2}{7}$

C

$\frac{3}{7}$

D

$\frac{4}{7}$

Solution

(a) Required probability $ = \frac{{{}^3{C_1}}}{{{}^7{C_1}}} \times \frac{{{}^2{C_1}}}{{{}^6{C_1}}} = \frac{1}{7}$.

Standard 11
Mathematics

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