**IIT-JEE Previous Years Solved Paper On Probability**

*Year-wise Solution of IIT-JEE paper on Probability*

**Q. [2010] A signal which can be green or red with probability 4/5 and 1/5 respectively,is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly is 3/4 .If the signal received at station B is green, then the probability that the original signal was green is:**

(a) 3/5 (b) 6/7 (c) 20/23 (d) 9/20

(a) 3/5 (b) 6/7 (c) 20/23 (d) 9/20

Sol:- (c)

Let us consider three events:

A₁ : The original signal is green.

A₂ : The original signal is red.

A : Signal received at station B is green.

P(A₁) = 4/5, P(A₂) =1/5,

P(A|A₁)=(3/4)² + (1/4)² =5/8,

P(A|A₂)=(3/4)(1/4)+(1/4)(3/4)=3/8

Now apply Bayes Theorem to find P(A₁|A). Ans is=20/23 .

**Q.[2008] An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. If A consists of 4 outcomes,the number of outcomes that B must have so that A and B are independent,is:**

(a) 2,4 or 8 (b) 3,6 or 9 (c) 4 or 8 (d) 5 or 10

(a) 2,4 or 8 (b) 3,6 or 9 (c) 4 or 8 (d) 5 or 10

Sol:- (d)

Given n(A)=4 ; P(A)=4/10

Let number of outcomes for B be n(B) ; P(B)=n(B)/10

Since A and B are independent, P(A∩B)=P(A)P(B)=(4/10)(n(B)/10)

⇒n(A∩B)=2(n(B))/5

Since n(A∩B) is an integer so n(B)=5 or 10.

**Q.[2004] If three distinct numbers are chosen randomly from the first 100 natural numbers,then the probability that all three of them are divisible by both 2 and 3 is:**

(a) 4/25 (b) 4/35 (c) 4/33 (d) 4/1155

(a) 4/25 (b) 4/35 (c) 4/33 (d) 4/1155

Sol:- (d)

A natural number is divisible by 2 or 3, if it is divisible by 6.

So, numbers divisible by 6 are 6,12,18, . . . . ,96

So there are total 16 numbers out of which 3 distinct number can be chosen in 16×15×14=n(A)

Total no of ways=100×99×98=n(S)

Required probability=n(A)/n(S)=4/1155 .

**Q.[2003] Two selected numbers are selected randomly from the set S={1,2,3,4,5,6} without replacement one by one. The probability that minimum of the two numbers is less than 4 is:**

(a) 1/15 (b) 14/15 (c) 1/5 (d) 4/5

(a) 1/15 (b) 14/15 (c) 1/5 (d) 4/5

Sol:- (d)

Total no of ways= 6C2=15.

Two numbers can be={(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)}

Number of favourable ways=12.

Required probability=12/15=4/5.

**Q.[1998] A fair coin is tossed repeatedly. If tail appears on the first four tosses,then the probability of head appearing in the fifth toss equals**

(a) 1/2 (b) 1/32 (c) 31/32 (d) 1/5

(a) 1/2 (b) 1/32 (c) 31/32 (d) 1/5

Sol:- (a)

Since the tosses are independent events. So, probability that a head appears on the 5th toss is not dependent on the earlier tosses. So, the required probability is 1/2.

__Note:- Each set on different topics will contain 5 distinct Solved problems from IIT-JEE papers. Keep visiting.__