A bag contains 12 white and 18 blacks balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?
Answer: D The probability that first ball is white= 12C1/30C1 = 12/30 =2/5. Since, the ball is not replaced, hence the number of balls left in bag is 29. Hence the probability the second ball is black =18C1/29C1 = 18/29. Required probability = 2/5 * 18/29 = 36/145
Q. No. 8:
Two brother X and Y appeared for an exam. The probability of selection of X is 1/7 and that of B is 2/9. Find the probability that both of them are selected.
Answer: C Let A be the event that X is selected and B is the event that Y is selected. P(A)= 1/7, P(B) = 2/9. Let C be the event that both are selected. P(C)= P(A)*P(B) as A and B are independent events = (1/7) * (2/9) = 2/63.
Q. No. 9:
In a race where 12 cars are running, the chance that car X will win is 1/6, that Y will win is 1/10 and that Z will win is 1/8. Assuming that a dead heat is impossible. Find the chance that one of them will win.
Answer: C P(P)= 1/4, P(Q)= 1/5, P (R)= 1/6 and P(S)= 1/7. All the events are mutually exclusive hence, Required probability = P(P)+P(Q)+P(R)+P(S) = 1/4 + 1/5 + 1/6 + 1/7 = 319/420
Q. No. 11:
From a bag containing 4 white and 5 black balls a man drawn 3 balls at random. What are the odds against these balls being black?
Answer: B Probability of all three balls being black = 5C3/9C3 = 5/42 Probability that three balls are not black = 1- 5/42 = 37/42 Hence, odds against these ball being black = 37/5 : 5/42 = 37 :5
Q. No. 12:
Dishant throws three dice in a special game. If it knows that he needs 15 or higher in this throw to win, then find the chance of his winning the game.