An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement, then the probability that both drawn balls are black, is

Question

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement, then the probability that both drawn balls are black, is (A) (2/7) (B) (1/7) (C) (5/7) (D) (3/7)

Tardigrade Admin · Accepted Answer

Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E ∩ F) or P(E F). Now, P(E)=P( black ball in first draw )=(10/15) Also, given that the first ball drawn is black, i.e., event F has occurred. Now, there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred. i.e., P(F / E)=(9/14) By multiplication rule of probability, we have P(E ∩ F) =P(E) ⋅ P(F / E) =(10/15) × (9/14)=(3/7)

Q. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement, then the probability that both drawn balls are black, is

$\frac{2}{7}$

$\frac{1}{7}$

$\frac{5}{7}$

$\frac{3}{7}$

Q. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement, then the probability that both drawn balls are black, is

72​

71​

75​

73​

$\frac{2}{7}$

$\frac{1}{7}$

$\frac{5}{7}$

$\frac{3}{7}$