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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