RBSE Solutions for Class 11 Maths Chapter 14 Probability Ex 14.2

Rajasthan Board RBSE Class 11 Maths Chapter 14 Probability Ex 14.2

Question 1.
A dice is thrown. Find the probability that number appear on dice is greater than 4.
Solution:
Total result obtained in throwing a dice.
S = {1,2, 3, 4, 5, 6}
then n(S) = 6
and n{E) = Total result to get digit more than 4 = 2
Thus, required probability
= \(\frac { n(E) }{ n(s) } \) = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)

Question 2.
A coin tossed two times. Find the probability that every time tail appear.
Solution:
in throwing two dice, total result obtained
S = {HH, HT, TH, TT}
Then favourable results n(E) = (1)
and n(S) = 4
Thus, required probability
P(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac { 1 }{ 4 } \)

Question 3.
From the natural numbers 1 to 17. One number is randomly selected. Find the probability that number is prime.
Solution:
Sample space of nos. 1 to 17.
S = {1,2, 3, 4,5,6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
E = (1, 3, 5, 7, 11, 13, 17}
Thus, n(S) = 17 and n(E) = 7
Thus required probability
P(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac { 7 }{ 17 } \)

Question 4.
A coin is tossed three times. find the probability that alternatively head of tail appear.
Solution:
In tossed a coin 3 times, sample space
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Result obtained alternatively
E = {HTH, THT)
Thus n(S) = 8 and n(E) = 2
Now required probability
P(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac { 2 }{ 8 } \) = \(\frac { 1 }{ 4 } \)

Question 5.
If two dices are thrown simultaneously, then find the probability that number appear are doublet or 2.
Solution:
If two dices are thrown simultaneously, then obtained sample space.
S = {(1, 1), (1, 2), (1,3), (1, 4), (1,5), (1,6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let E₁ = Prob. to get doublet
and E₂ = Prob. to get sum 9
There E₁ = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
n(E₁) = 6
E₂ = {(3, 6), (4, 5), (5, 4), (6, 3)}
n(E₂) = 4

Question 6.
Find the probability that there are 52 Sundays in a normal (not leap) year.
Solution:
There are 365 days in a non-leap year 365 – 52 = 7 and 1 day. Remaining 1 days may be any day from Monday to Sunday = \(\frac { 1 }{ 7 } \)
Thus probability of being Sunday
= 1 – \(\frac { 1 }{ 7 } \) = \(\frac { 6 }{ 7 } \)
Here we should take probability of not being Sunday here we should take probability of not being Sunday
∴ Required Probability = \(\frac { 6 }{ 7 } \)

Question 7.
A card from a deck of 52 cards is draw, find the odds in the favour of ace cad.
Solution:
Total result n(S) = 52
Total results of card being are
n(E) = 4
Thus Probability to get are

Question 8.
In a class of 12 students there are 5 boys and remainings are girls. In the selection of a student find the odds against selection of girl.
Solution:
Total students = 12
∴ n(S)= 12
Total boys n(E₁) = 5
and Total girts = n(E₂) = 12 – 5 = 7
Probability to select girl

Question 9.
Persons are sitting arouund a table. If two specific persons sit simultaneously. What will be the odds in this position ?
Solution:
Let any person occupy any seat then (n – 1) seats are available for other specific person. If both sit simultaneously, then their probability

Question 10.
There are three letters and three corresponding envelop. If each of all the letters are kept randomly in envelop, then what will be the probability that all the letters are kept in right envelops ?
Solution:
Total ways to kept 3 letters in3 envelops
∴ Pro. of kept the letter in right way = \(\frac { 1 }{ 31 } \)
= \(\frac { 1 }{ 3\times 2 } \) = \(\frac { 1 }{ 6 } \)
Thus Prob. of kept the letter in wrong way = 1 – \(\frac { 1 }{ 6 } \) = \(\frac { 5 }{ 6 } \)

Question 11.
Out of first two hundred intgers, one digit is randomly chosen.Find the probability that it will divide by 6, 8, or 24.
Solution:
Total no. of integers = 100
∴ n (S) = 100
Favourable cases = (Divisible by 6) + (Divisible by 8) – (Deivisible by 24) = {6, 12, 18, …, 198}
+ {8, 16, 24, …,200} – {24, 48, …, 192}
by formula l = a(n – 1 )d
= 33 + 25 – 8 = 50
∴ Required Prob. = \(\frac { 50 }{ 200 } \) = \(\frac { 1 }{ 4 } \)

Question 12.
If three dices are thrown simultaneously. Find the probability the sum of digits obtained is more than 15.
Solution:
In throw of three dices, sample space
S = {(3, 6, 6), (4, 5, 6), (4, 6, 5), (5, 5, 5), (5, 6, 4), (5, 4, 6), (6, 3, 6), (6, 6, 3), (6, 5, 4), (6, 4, 5)}

Question 13.
Letters of word ANGLE are arranged in a row randomly. Find the probability the vowels occurs together.
Solution:
By arranging five lettters of word ANGLE
= 5 ! = 5 × 4 × 3 × 2 = 120
All vowels ocurs together,then
Fovourable cases = 4 ! × 2 !
= 4 × 3 × 2 × 2 = 48
∴ Required Probability = \(\frac { 48 }{ 120 } \) = \(\frac { 2 }{ 5 } \)

Question 14.
A card is drawn from a deck of card.Find the probability that chosen card is ace, king or queen.
Solution:
52 prob. to get one ace out of 52 cards

Prob. to get one king out of 52 cards

Similarly, prob. to get one queen out of 52 cards

Thus, probability to get ace, king or queen.

Question 15.
A bag contains 6 white, 7 red and 5 black balls. Out of these 3 balls are ramdomly chosen one by one.What will be the probability that three balls are white,whereas the ball drawn is not replaced back ?
Solution:
Probability to draw 1 ball out of 6 white balls whereas total balls are 18

In second attempt
Probability to draw 1 ball out of 5 white ball whereas total balls are 17

In third attempt
Probability to draw 1 ball out of 4 white balls whereas total balls are 16.

Thus, total probability that all the three balls are while.

RBSE Solutions for Class 11 Maths