Q: In a binomial distribution, the mean is 4 and variance is 3 . Then, its mode is

aligned Mean & =n p=4 \\ Variance & =n p q=3 \\ q & = 3 4 aligned and p=1-q=1- 3 4 = 1 4 n=16 Now, Mode of Binomial distribution is given by (n+1) P Case IIf (n+1) P= Integer (I), then Mode =\ I, I-1\ Case II If (n+1) P Integer (I+f), then Mode =\ I\ aligned (n+1) P & =(16+1) 1 4 \\ & = 17 4 =4.25 \\ & =4+0.25 \\ Mode & =4 aligned

Q: If E and F are events such that 0 < P ( F ) < 1, then

P(E F)+P( E F) = P(E F)+P( E F) P(F) = P((E E ) F) P(F) = P(F) P(F) =1

Q: The probability of getting sum more than 7 when a pair of dice are thrown is:

Here n(S)=6^2=36 Let E be the event "getting sum more than 7" i.e. sum of pair of dice =8,9,10,11,12 i.e. E= \ array lllll (2,6) & (3,5) & (4,4) & (5,3) & (6,2) \\ (3,6) & (4,5) & (5,4) & (6,3) \\ (4,6) & (5,5) & (6,4) & \\ (5,6) & (6,5) & (6,6) & array \ n(E)=15 . Req. prob = n(E) n(S) = 15 36 = 5 12

Question 1

The probability of getting 10 in a single throw of three fair dice is

Accepted Answer

Total outcomes =216 Now, number of cases of getting 10 from 3 dices in single throw are Case 1:~1+3+6 outcomes =3!=6 Case 2:~1+4+5 outcomes =3 ! =6 Case 3:~2+2+6 outcomes = 3! 2! =3 Case 4:~2+3+5 outcomes =3!=6 Case 5:~2+4+4 outcomes = 3! 2! =3 Case 6:~3+3+4 outcomes = 3! 2! =3 Favourable outcomes =27 Probability = 27 216 = 1 8 aligned & & (x+24)(x-20) & =0 \ & & x & =20 aligned

Question 2

In a binomial distribution, the mean is 4 and variance is 3 . Then, its mode is

Accepted Answer

aligned Mean & =n p=4 \ Variance & =n p q=3 \ q & = 3 4 aligned and p=1-q=1- 3 4 = 1 4 n=16 Now, Mode of Binomial distribution is given by (n+1) P Case IIf (n+1) P= Integer (I), then Mode =\ I, I-1\ Case II If (n+1) P Integer (I+f), then Mode =\ I\ aligned (n+1) P & =(16+1) 1 4 \ & = 17 4 =4.25 \ & =4+0.25 \ Mode & =4 aligned

Question 3

The probability that certain electronic component fails when first used is 0.10 . If it does not fail immediately, the probability that it lasts for one year is 0.99 . The probability that a new compo

Accepted Answer

P (The electronic component fails when first aligned & used )=P(F)=0.10 \ & P( F )=1-P(F)=0.90 aligned Let E be the event that a new component will last for one year, then P(E)=P(F) P ( E F )+P( F ) P ( E F ) [Total probability theorem] =0.10 0+0.90 0.99=0.891

Question 4

Given below is the distribution of a random variable X If = P (X 3) and = P (X 2), then : =

Accepted Answer

For a distribution of random variable x, =P (X^6 3 )=P (X^6=1 )+P (X^6=2 ) aligned & = +2 =3 \ & and =P (X^6 2 )=P (X^6=3 )+P (X^6=4 ) \ & =3 +4 =3 \ & : =3: 7 aligned

Question 5

A book contains 1000 pages. A page is chosen at random. The probability that the sum of the digits of the marked number on the page is equal to 9 , is

Accepted Answer

n(S)= Total number of ways =1000 The favourable cases that the sum of the digits of the marked number on the page is equal to 9 are one digit number or two digits numbers or three digits numbers, if three digit number is a b c. Then, a+b+c=9 ; 0 a, b, c 9. n(E)= Number of favourable ways = Number of solutions of the equation = ^ 9+3-1 C_ 3-1 = ^ 11 C_2=55 Required probability = n(E) n(S) = 55 1000 = 11 200 .

Question 6

For two events A and B , if P ( A )= P ( A B )= 1 4 and P ( B A )= 1 2 , then which of the following is not true?

Accepted Answer

P ( B A )= 1 2 P ( B A ) P ( A ) = 1 2 P ( B A )= 1 8 P ( A B )= 1 4 P ( A B ) P ( B ) = 1 4 P ( B )= 1 2 P ( A B )= 1 8 = P ( A ) P ( B ) Events A and B are independent. Now, P ( A ^ B )= P ( A B ) P ( B ) = P ( A ) P ( B ) P ( B ) = 3 4 and P ( B ^ A ^ )= P ( B ^ A ^ ) P ( A ^ ) = P ( B ^ ) P ( A ^ ) P ( A ^ ) = 1 2

Question 7

If A and B are mutually exclusive events and if P ( B )= 1 3 , P ( A B )= 13 21 , then P ( A ) is equal to

Accepted Answer

For mutually exclusive events P ( A B )= P ( A )+ P ( B ) P ( A )= 2 7

Question 8

A coin is tossed twice. Then, the probability that atleast one tail occurs is

Accepted Answer

The sample space is S =\ HH , HT , TH , TT \ Let E be the event of getting atleast one tail aligned & E =\ HT , TH , TT \ \ & Required probability p \ & = Number of favourable outcomes Total number of outcomes = n ( E ) n ( S ) = 3 4 aligned

Question 9

If E and F are events such that 0 < P ( F ) < 1, then

Accepted Answer

P(E F)+P( E F) = P(E F)+P( E F) P(F) = P((E E ) F) P(F) = P(F) P(F) =1

Question 10

The probability of getting sum more than 7 when a pair of dice are thrown is:

Accepted Answer

Here n(S)=6^2=36 Let E be the event "getting sum more than 7" i.e. sum of pair of dice =8,9,10,11,12 i.e. E= \ array lllll (2,6) & (3,5) & (4,4) & (5,3) & (6,2) \ (3,6) & (4,5) & (5,4) & (6,3) \ (4,6) & (5,5) & (6,4) & \ (5,6) & (6,5) & (6,6) & array \ n(E)=15 . Req. prob = n(E) n(S) = 15 36 = 5 12

Question 11

The probability that a card drawn from a pack of 52 cards will be a diamond or king is:

Accepted Answer

Total no. of cards =52 13 cards are diamonds and 4 cards are king. There is only one card which is a king of diamond. P ( card is diamond )= 13 52 , P (card is king) = 4 52 P ( card is king of diamond )= 1 52 P (card is diamond or king) = 13 52 + 4 52 - 1 52 = 16 52 = 4 13

Question 12

Bag P contains 6 red and 4 blue balls and bag Q contains 5 red and 6 blue balls. A ball is transferred from bag P to bag Q and then a ball is drawn from bag Q. What is the probability that the ball dr

Accepted Answer

Let E_1, E_2 and A be the events defined as follows: E _1= red ball is transferred from bag P to bag Q E _2= blue ball is transferred from bag P to bag Q A = the ball drawn from bag Q is blue As the bag P contains 6 red and 4 blue balls, P (E_1 )= 6 10 = 3 5 and P (E_2 )= 4 10 = 2 5 Note that E_1 and E_2 are mutually exclusive and exhaustive events. When E_1 has occurred i.e., a red ball has already been transferred from bag P to Q, then bag Q will contain 6 red and 6 blue balls, So, P (A E_1 )=

Question 13

If P ( B )= 3 5 , P ( A B )= 1 2 and P ( A B )= 4 5 , then P ( A B )^ + P ( A ^ B )=

Accepted Answer

P(B)= 3 5 , P(A B)= 1 2 and P(A B)= 4 5 P(A B)=P(A B) P(B)= 1 2 3 5 = 3 10 P ( A B )= P ( A )+ P ( B )- P ( A B ) P ( A )= 4 5 - 3 10 = 1 2 . P ( A ^ )=1- P ( A )= 1 2 We know, P ( A B )+ P ( A ^ B )= P ( B ) [as A B and A^ B are mutually exclusive events] 3 10 + P ( A ^ B )= 3 5 P ( A ^ B )= 3 5 - 3 10 = 3 10 Now, P ( A ^ B )= P ( A ^ )+ P ( B )- P ( A ^ B ) = 1 2 + 3 5 - 3 10 = 5+6-3 10 = 4 5 P (( A B )^ )=1- P ( A B )=1- 4 5 = 1 5 P (( A B )^ )+ P ( A ^ B )= 1 5 + 4 5 =1

Question 14

A six faced die is a biased one. It is thrice more likely to show an odd numbers than show an even number. It is thrown twice. The probability that the sum of the numbers in two throws is even, is

Accepted Answer

Let P be the probability of getting an even number. Then by hypothesis the probability of getting an odd number is 3 P. Since the events of getting an even number and an odd number are mutually exclusive and exhaustive. P+3 P=1 P=1 / 4 Thus, the probability of getting an odd number in a single throw is 3 4 and that of an even number is 1 4 . If the die is thrown twice, then the sum of the numbers in two throws is even if both the numbers are even or both are odd. Required probability = 3 4 3 4 +

Question 15

Girl students constitute 10 \% of I year and 5 \% of II year at Roorkee University. During summer holidays 70 \% of the I year and 30 \% of II year students are given a project. The girls take turns o

Accepted Answer

The desired probability = 10 100 70 100 10 100 70 100 + 5 100 30 100 = 14 17

Question 16

The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then P ( X =1) is

Accepted Answer

aligned n p &=4 \ n p q &=2 \ q &= 1 2 , p= 1 2 , n=8 \ P ( X &=1)= ^ 8 C _ 1 ( 1 2 ) ( 1 2 )^ 7 =8 1 2^ 8 = 1 2^ 5 = 1 32 aligned

Question 17

The probability that certain electronic component fails when first used is 0.10 . If it does not fail immediately, the probability that it lasts for one year is 0.99 . The probability that a new compo

Accepted Answer

Probability that the electronic component fails when first used is P(F)=0.10. Therefore, P (F^ )=1-P(F)=0.90 Let E be the event that a new component will last for one year. Then, P(E)=P(F) P ( E F )+P (F^ ) P ( E F^ ) [total probability theorem] =0.10 0+0.90 0.99=0.891

Probability — Previous Year Questions & Solutions