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(FF). Now, P(E) = P(black ball in first draw) =(10/15) 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​

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(FF). Now, P(E)=P(black ball in first draw) =1510​ P(F∣E)=149​ By multiplication rule of probability, we have P(E∩F)=P(E).P(F∣E) =1510​×149​=73​

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

$\frac{2}{7}$

$\frac{1}{7}$

$\frac{5}{7}$

$\frac{3}{7}$