Four different objects 1,2,3,4 are distributed at random in four places marked 1,2,3,4. What is the probability that none of the objects occupy the place corresponding to its number?

Answer: C Let a particular number (say) number 2 occupies position 1. Then all possible arrangement are given (2,1,3,4), (2,1,4,3), (2,3,4,1), (2,4,1,4), (2,4,1,3), (2,4,3,1). Out of these six three (2,1,3,4), (2,3,1,4), (2,4,3,1) are not acceptable because numbers 3 and 4 occupy the correct positions. Required probability = 3/6 = 1/2

Q. No. 2:

There are 6 positive and 8 negative numbers. Four numbers are chosen at random and multiplied. The probability that the product is a positive number is

Answer: C The product of four numbers will be positive in the following ways. 1). All the four numbers are positive, hence probability =6_{C4}/14_{C4} 2).All the four numbers are negative, hence probability =8_{C4}/14_{C4} 3). Two numbers are positive and two are negative, hence Probability = 6_{C2}*8_{C2}/14_{C4} Hence, required probability of the event = (6_{C4}+8_{C4}+6_{C2}*8_{C2})/ 14_{C4} = (15+70+15*28)/1001 = 505/1001

Q. No. 3:

A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A bag and a ball are picked random. The probability that the ball will be white is

Answer: D The probability of selecting one bag = 1/2 Now, probability of getting a white ball from bag A = 1/2 * 3/5 = 3/10 and probability of getting a white ball from bag B = 1/2 * 2/6 = 1/6 Hence, Probability that white ball is drawn either first or second bag = 3/10 + 1/6 = 7/15

Q. No. 4:

One hundred identical coins each with probability 'p' showing up heads and tossed. If 0<p<1 and the probability of heads showing on 50 coins is equal to that of heads on 51 coins, then the value of p is

Answer: D Let a be the number of coins showing heads then, P(A=50) = P(A=51) =>100_{C50}*P^{50}*(1-P)^{50} => 100_{C51}*P^{51}*(1-P)^{49} => 100_{C50}*(1-P) = 100_{C51}*P => 51(1-P)= 50P P= 51/101

Q. No. 5:

The probability that a student is not a swimmer is 1/5. Then the probability that one of the five students, four are swimmers is

Answer: C 4 students out of 5 can be selected in
5_{C4} ways.Probability of a student being not a swimmer = 1/5 Probability of a student being a swimmer = (1- 1/5) = 4/5 Required probability =5_{C4}*(1/5)*(4/5)^{4}

Q. No. 6:

If the probability that X will live 15 year is 7/8 and that Y will live 15 years is 9/10, then what is the probability that both will live after 15 years?