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Solved Objective Question on Probability Set 2

Posted on - 21-05-2017

JEE Math Probability

IIT JEE

Question.1

A fair die is thrown until a score of less than 5 points is obtained. The probability of obtaining not less than 2 points on the last thrown is.

(A) 3/4

(B) 5/6

(C) 4/5

(D) 1/3

Solution

Score less than 5 means the occurrence of 1, 2, 3, or 4. Now on the last throw we should not obtain a score less then 2 i.e. one. Clearly the favourable outcomes are 2, 3 or 4. .

Thus the required probability = 3/4

Question.2

Three persons A1, A2 and A3 are to speak at a function along with 5 other persons. If the person speak in random order, the probability that A1 speaks before A2 and A2 speaks before A3 is’.

(A) 1/6

(B) 3/5

(C) 3/8

(D) none of these

Solution

Total number of ways in which 8 persons can speak is 8!

Now 3 positions out of 8 position can be chosen in 8C3 i.e. 56 ways and at these positions we can put A1, A2 and A3 in the required order. Further the remaining persons can speak in 5! ways

&⇒
Total number of favourable ways
= 56 (5! )
&⇒
Required probability = = 1/6.

Question.3

A fair die is tossed eight times. Probability that on the eighth throw a third six is observed is, .

(A) 8C3

(B)
(C)

(D) none of these

Solution

Third six occurs on 8th trial. It means that in first 7 trials we must exactly 2 sixes and 8th trial must result in a six.

&⇒
Required probability = 7C2 . (1/6)2. (5/6)5. (1/6)
= .

Question.4

A fair coin is tossed a fixed number of times . If the probability of getting 7 heads is equal to getting 9 heads, then the probability of getting 2 heads is,
(A) 15/28 .

(B) 2/15

(C) 15/213

(D) none of these

Solution

Let coin was tossed 'n' times and X be the random variable representing the number of head appearing in 'n' trials.
P(X = 7) = P (X = 9) .


&⇒
n C7(1/2)7 (1/2)n-7 = nC9(1/2)n-9 ´ (1/2)9

&⇒
nC7 = nC9
&⇒
n = 16

Now P (X = 2) = 16C2(1/2)2 (1/2)14

= = .

Question.5

If the papers of 4 students can be checked by any one of the 7 teachers, then the probability that all the 4papers are checked by exactly 2 teachers is;

(A) 2/ 7

(B) 32/ 343

(B) 12/ 49

(D) None of these

Solution

Total number of ways in which 4 papers can be distributed among 7 teachers = 74 .

Now exactly 2 teachers out of 7 can be chosen in 7C2 ways. And total number of ways in which 4 papers can be given to these 2 teachers ( each one getting atleast one) = (24 –2 ) =14.


&⇒
Total number of ways in which exactly 2 teachers check all four papers = 7C2
. 14= 21 .14.


&⇒
Required probability =

Question.6

Let 'E' and 'F' be two independent events. The probability that both 'E’ and 'F’ happen is 1/12 and the probability that neither 'E' nor 'F' happens is ½, then ,.

(A) P(E) = 1/3, P(F) = 1/4

(B) P(E) = 1/2, P(F) = 1/6

(C) P(E) = 1/6, P(F) = 1/2

(D) P(E) = 1/4, P(F) = 1/3

Solution

P(EÇF) = P(E).P(F) = 1/12.

P(E¢ÇF¢) = P(E¢).P(F¢) .


&⇒
1/2 = ( 1 – P(E) ) ( 1- P(F))
&⇒
P( E)
.P(F) + 1 – P(E) – P(F) = 1/2.


&⇒
P(E) + P(F) =
&⇒
P(E) = 1/3, P(F) = 1/4

or P(E) = 1/4, P(F) = 1/3

Question.7

There are n persons (n ≥ 3), among whom are A and B, who are made to stand in a row in random order. Probability that there is exactly one person between A and B is .

(A)

(B)

(C) 2/n

(D) none of these

Solution

Person that must stand between A and B can be chosen in (n-2) ways. Now number of ways in which x person can be made stand so that there is exactly one person in between A and B is equal to (n-2) . 2. (n-2)! .

Also total number of ways in which persons can be made to stand =n!

&⇒
Required probability = =

Question.8

A number is chosen at random from the numbers 10 to 99. By seeing the number a man will laugh if product of the digits is 12. If he choose three numbers with replacement then the probability that he will laugh at least once is .

(A) 1 –

(B)

(C) 1 –

(D) 1 –

Solution

There can be four such numbers i.e. 43, 34, 62, 26.

Whose product of digit is 12.


&⇒
Probability that the man will laugh by seeing the chosen numbers

=


&⇒
Required probability = 1– .

Question.9

In a bag there are 15 red and 5 white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is .

(A) 12/19

(B) 13/19

(C) 14/19

(D) 15/19

Solution

Probability that out of remaining balls the one that is red is

=

Question.10

If ‘head’ means one and ‘tail’ means two , then coefficient of quadratic equation ax2 + bx + c = 0 are chosen by tossing three fair coins. The probability that roots of the equations are imaginary is .

(A) 5/8

(B) 3/8

(C) 7/8

(D) 1/8

Solution

b2 – 4ac < 0

For b =1any a and c which can be chosen in 4 ways

For b =2 either a = 1, c = 2

or a = 2, c =1

or a = 2, c = 2


&⇒
Required probability = 7/8

 
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