14.Probability
medium

दो गेंद एक बॉक्स से बिना प्रतिस्थापित किए निकाली जाती है। बॉक्स में $10$ काली और $8$ लाल गेदें हैं तो प्रायिकता ज्ञात कीजिए एक काली तथा दूसरी लाल हो।

A

$\frac{40}{81}$

B

$\frac{40}{81}$

C

$\frac{40}{81}$

D

$\frac{40}{81}$

Solution

Total number of balls $=18$

Number of red balls $=8$

Number of black balls $=10$

Probability of getting first ball as red $=\frac{8}{18}=\frac{4}{9}$

The ball is replaced after the first draw.

Probability of getting second ball as black $=\frac{10}{18}=\frac{5}{9}$

Therefore, probability of getting first ball as black and second ball as red $=\frac{4}{9} \times \frac{5}{9}=\frac{20}{81}$

Therefore, probability that one of them is black and other is red

$=$ Probability of getting first ball black and second as red $+$ Probability of getting first ball red and second ball black

$=\frac{20}{81}+\frac{20}{81}$

$=\frac{40}{81}$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.