RBSE Solutions for Class 11 Maths Chapter 14 Probability

RBSE Solutions for Class 11 Maths Chapter 14 Probability

Rajasthan Board RBSE Class 11 Maths Chapter 14 Probability Ex 14.1

Question 1.
From a cartoon of bulbs, 3 bulbs are drawn randomly. Test each bulb and classified into damage (D) and not damage (N). Find the sample space of this test.
Solution:
After drawing first bulb two results D of will be
After drawning 2nd bulb result
= (D,N) × (D,N)
= {DD, DN, ND, NN}
After drawing 3rd bulb result
= (D, N) × {DD, ND, NN}
= {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}
Thus sample space
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}

Question 2.
Four cards are drawn from a desk of cards. What will be the n (E) whereas E is an event to draw 1 king 1 queen 1 fack 1 ace.
Solution:
n(E) = Total ways to get a queen
× Total ways to get a queen
× Total ways to get a fack
× Total ways to get 1 ace
= 4 × 4 × 4 × 4 = 256.

Question 3.
A dice is thrown E is an event to getting 4 on dice and getting even number is event F. Are the events E and F mutually exclusive events ?
Solution:
In throwing a dice, sample space
S = {1, 2, 3, 4, 5, 6}
Then according to question
E = {4}
and F = {2, 4, 6}
Thus E ∩ F = {4} ≠ Φ
Thus E and F are not mutually exclusive events.

Question 4.
Two dice and thrown, then
(i) What is the sample space for pair ?
(ii) What is the sample space for sum of digits appear is 8 ?
Solution:
(i) For pair, sample space.
= {(1,1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
(ii) For sum of digit 8, sample space.
= {(2, 6), (3, 5), (4, 4), (5, 3), (6, 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
= n(E)/n(s) = 2/6 = 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) = n(E)/n(S) = 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) = n(E)/n(S) = 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) = n(E)/n(S) = 2/8 = 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 = 1/7
Thus probability of being Sunday
= 1 – 1/7 = 6/7
Here we should take probability of not being Sunday here we should take probability of not being Sunday
∴ Required Probability = 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 = 1/31
= 1/3×2 = 1/6
Thus Prob. of kept the letter in wrong way = 1 – 1/6 = 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. = 50/200 = 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 = 48/120 = 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.

Rajasthan Board RBSE Class 11 Maths Chapter 14 Probability Ex 14.3

Question 1.
Probability of event A is 2/11, find the probability of event ‘A-not’.
Solution:

Question 2.
There are 4 male and 6 female members in village paneLIf one member for a committee is selected randomly then what is the probability to select a female.
Solution:
Required probability

Thus, probability, to select a female is 3/5

Question 3.
In throwing a dice find the probability of following events.
(i) Appear a prime number
(ii) apear 1 or less than 1
(iii) appear number less than 6.
Solution:
In throwing a dice, obtained sample spapce
S = {1,2, 3, 4, 5, 6}, n(S) = 6
(i) Event E, appear a prime number
then E = {2, 3, 5}
Thus n(E) = 3
Then probability of event E

Thus, probability to get a prime number is = 1/2
(ii) Event to get number 1 or less than 1
B = {1}
Then n(B) = 1
Thus, probability to get 1 or less than 1

Thus, probability to get no. 1 or less than 1 = 1/6
(iii) Event to appear no. less than 6
D = {1,2, 3,4,5}
then n(D) = 5
Then, probability to appear no. less than 6.

Thus, probability to get number less than 6 = 5/6

Question 4.
A coin is tossed 4 times. Find the probability to get tail at least three times in these throw.
Solution:
Sample space obtained by throwing a coin four time
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTP, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
n(S) = 16
Let appearance of tail at least 3 times is event A, then
A = {HHHH, HHHT, HHTH, HTHT, HTHH}
n(A) = 5
Thus, required probability

Thus, probability to get tail at least 3 times is 5/16

Question 5.
If a coin and a dice are thrown simultaneously, then what is the probability that tail appear on coin and event number appear on dice ?
Solution:
In throwing coin and dice obtained sample space
S = {H, T} × {1, 2, 3, 4, 5, 6}
= {H1, H2, H3, H4, H5, H6, T1, T2, 73, T4, T5, T6}
n(S) = 12
Let A be the sample space of getting tail on coin and event number on dice, then
A = {H2, H4, H6}
n(A) = 3
Thus, required probability

Thus, required probability is 1/4.

Question 6.
In a company of 20 people 5 are graduate. If 3 people are selected randomly then what is the probability that one of them be graduate ?
Solution:
Total number of ways in which 3 person are to be select out of 20 = ²⁰C₃
Total number of ways to select one graduate = ⁵C₂
and not to select any graduate = ¹⁵C₂
∴ Total favaurable situations = ¹⁵C₂ × ⁵C₁
Hence, required probability

Question 7.
To solve the problem in opposite of A, odds are 4 : 3. In favour of B odds are 4 : 3. What is the probability that –
(i) Problem wil be solved
(ii) Problem will not be solved
(iii) Problem will be solved only by one
Solution:
Given

(i) Probability of problem solved

(ii) Prob. for problem not solved

(iii) Probability of to be solved by only one

Question 8.
An equipment will work only when its three components A, B and C are in working condition. Probability that A will not work is 0.15 and (M)5 of B and 0.10 to C. What is the probability that equipment will not be in working condition before year ending ?
Solution:
According to question

Probability of damage equipment before year ending
= 1 -P(A).P(B).P(C)
= 1 -0.72675 = 0.27325

Question 9.
From a deck of cards,two cards are randomly drawn one by one. If card once drawn is not replaced then to get two aces in first attempt and two king in second attempt. What is the probability ?
Solution:
Total cards in deck = 52
Probability to get 2 aces in 1st attempt

Probability to get 2 kings in IInd attempt

Question 10.
A and B are two events in which P(A) = 1/3. P(B) = 1/4 and P(AB) = 1/12 then find P(B/A)
Solution:

Question 11.
Imagine that ratio of male and children are 1:2. Find the probability that in a family out of 5 children are (i) boys (ii) three boys and two girls.
Solution:
Let number of males = x
and Number of boys = x
and Number of girls = x
Probability of choosing boys (P)

(i) Probability of being all boys out of 5 children

(ii) Probability of being 3 boys and 2 girls out of 5 children

Question 12.
A can hit the target 3 times out of 6, 8 can 2 times out of 4 and C can one times out of 5. They let the target simulatenously. What is the probability that at least two persons can hit the target.
Solution:
Given

Probability that at least two person hit the target

Rajasthan Board RBSE Class 11 Maths Chapter 14 Probability Miscellaneous Exercise

Question 1.
In throwing a coin n times, n(S) is –
(A) 2n
(B) 2ⁿ
(C) n²
(D) n/2
Solution:
Thus (B) is correct.

Question 2.
In throwing two dice and sample space of getting sum 3 is –
(A) (1, 2)
(B) {(2, 1)}
(C) {(3, 3)}
(D) {(1, 2), (2, 1)}
Solution:
Thus (D) is correct.

Question 3.
If tossing a coin and a dice simultaneously the number of elements in sample space is—
(A) 12
(B) 6
(C) 64
(D) 36
Solution:
No. of elements = n{H, T} × n{1, 2,3,4,5,6}
= 2 × 6 = 12
Thus, (A) is correct.

Question 4.
Result of each experiment is called –
(A) Sample space
(B) Random test
(C) Sample point
(D) Ordered pair
Solution:
Thus, (C) is correct.

Question 5.
If three coin are tossed and E be the event to getting at least one head, then n(E) will be –
(A) 6
(B) 3
(C) 4
(D) 8
Solution:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, ITT)
n(E) = 4
Thus, (C) is correct.

Question 6.
if E₁ ∩ E₂ = Φ, then E₁ and E₂ will be –
(A) Exclusive
(B) Independent
(C) Dependent
(D) Complementary
Solution:
Thus, (D) is correct.

Question 7.
Favourate events of 53 Mondays in leap year will be –
(A) 7
(B) 2
(C) 1
(D) 14
Solution:
Total days in a leap year = 366
Remaining day of after 53 weeks
= 366/53 = 2 remaining
Thus, (B) is correct.

Question 8.
Three are 4 white 5 black and 2 red balls in an urn. Favourable cases of three different colour balls will be –
(A) 9
(B) 24
(C) 12
(D) 7
Solution:
White balls = 4, Black balls = 3
and Red balls = 2
Favourable cases of three different colour balls
= 4 × 3 × 2 = 24
Thus, (B) is correct.

Question 9.
In two mutually exclusive events value of P (A ∪ B) is –
(A) P(A) + P(B)
(B) P(A) + P(B) – P(A ∩ B)
(C) P(A).P(B)
(D) P(A).P(B/A)
Solution:
Thus, (A) is correct.

Question 10.
The probability of solving the question by three students A,B and C are 1/2, 1/3 and 1/4, then probability of solving the question by at least one is –
(A) 1/24
(B) 1/4
(C) 3/4
(D) 1/9
Solution:
P(A)= 1/2, P(B) = 1/3, P(C) = 1/4
Required probability

Thus, option (C) is correct.

Question 11.
On tossing two dices simultaneously, probability to getting diference of numbers appear as 1 will be –
(A) 5/18
(B) 1/4
(C) 2/9
(D) 7/36
Solution:
Total events = 6 × 6 = 36 ⇒ n(S) = 36
Favourable events
A = {(1, 2), (3, 4), (4, 5), (5, 6), (2, 3), (6, 5), (5, 4), (4, 3), (3, 2), (2, 1)}
n(A) = 10
Thus, Required probability

Thus, (A) is correct.

Question 12.
A card is drawn from a deck of cards, probability of getting red or black card is –
(A) 1/4
(B) 1/2
(C) 3/4
(D) 26/51
Solution:
Required probability

Thus, (B) is correct.

Question 13.
On throwing two dices probability to getting sum of numbers appear as multiple of 4 will be –
(A) 1/4
(B) 1/3
(C) 1/9
(D) 5/9
Solution:
Multiples of 4 are = 4, 8, 12, 16,20,24, 28, 32, 36, 40, … etc.
Thus favourable events A = {(1, 3), (3, 1), (4,4), (5, 3), (3,5), (6, 2), (6,6)}
n(A) = 9 and n(S) = 36
Thus, required probability

Thus, (A) is correct.

Question 14.
If 5 digit numbers are formed by using digit 1,2,3,4, 5,6 and 8, then probability to get even digit at both end will be –
(A) 5/7
(B) 4/7
(C) 3/7
(D) 2/7
Solution:
No. of total digits = 7
Even number appear on both end so, favourable cases = 2
Thus, required probability = 2/7
Thus, (D) is correct.

Question 15.
In throwing three dices probability to get same digit on all three is –
(A) 1/36
(B) 3/22
(C) 1/6
(D) 1/18
Solution:
∵ A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6)}
n(A) = 6, n(S) = 6³ = 216
Required probability

Thus, (A) is correct.

Question 16.
In a swimming race odds in favour of A is 2 : 3 and odds in opposite of B is 4 : 1. Find the probability of winning of A orB is –
(A) 1/5
(B) 2/5
(C) 3/5
(D) 4/5
Solution:

Question 17.
10 students are sitting in a raw randomly, probability that two specific students do not sit closely is –
(A) 1/5
(B) 2/5
(C) 3/5
(D) 4/5
Solution:
Required probability

Thus, (D) is correct.

Question 18.
There are 12 sections in a group, in which four sections are faulty 3 sections are randomly drawn one by one. Without replacement probability to get no. one as faulty is –
(A) 3/55
(B) 13/55
(C) 14/55
(D) 17/55
Solution :
Total terms = 12, Non-facility sections = 12 -4 = 8
Faulty section = 4
Probability to get no. one as fault in three attempt is –

Thus (C) is correct.

Question 19.
Probability of any sure event will be –
(A) 0
(B) 1/2
(C) 1
(D) 2
Solution:
Thus (C) is correct.

Question 20.
A family having 3 children in which at least one boy, then probability that family have one boy and one girl is –
(A) 1/2
(B) 1/3
(C) 1/4
(D) 3/4
Solution:
Sample space of children in a family is :
S = {B, G, G, B, B, G, B, B, G}
Event of 2 boys and 1 girl A = {B, B, G}
∴ n(S) = 3, n(A) = 1
Thus, required probability = n(A)/n(S) = 1/3
Thus, (B) is correct.

Question 21.
The probability of taking examination in class by a teacher is 1/5. If a student is absent two times then probability that he was absent at least in one examination is –
(A) 9/25
(B) 11/25
(C) 13/25
(D) 23/25
Solution:
Let student is absent in first examination, event is collect E and absent in 2nd is F, then
According to question

Probability that students is absent in at least one exam. =
1 – (Prob. that student is absent in both the exams)

Thus, this is prob. that student appear at least one examination.
Thus, (A) is correct.

Question 22.
In a non-leap year. Find the probability to getting 53 Mondays.
Solution:
There are 52 Sundays in a leap year and 2 days are remains. If one of them is Sunday, then there will be 53 Sundays.
Now, probability that at least two days one of them is Sunday = 1/7
∴ Required probability = 1/7

Question 23.
A and B are two mutually exclusive events and P(A) – 0.3, P(B) = K and P(A ∪B) = 0.5, then find the value of K –
Solution:
P(A ∪ B) = P(A) + P(B)
⇒ P{B) = P(A ∪ B) – P(A)
⇒ P(B) = K = 0.5 – 0.3
= 0.2
Thus K = 0.2.

Question 24.
Words are formed by using letters of word ‘PEACE’. Find the probability that word contains both E.
Solution:
Total words formed by letters of word ‘PEACE’

Question 25.
There are 6 red and 8 black balls in a bag. 4 balls are taken out two times, 4 balls taken once are replaced back. What will be the probability that 4 balls in first attempt be red and in second attempt be black ?
Solution:
Total balls = 6 + 8 = 14
Since ball are replaced
∴ Probability to draw 4 red ball in 1st chance and 4 black balls in 2nd chance.

Question 26.
A man speaks truth 3 times out of 5. He says that in tossing 6 coins,two tails appear so what is the probability that this event is actually true ?
Solution:
Let E represents the statement of a person. Now, S₁ is the event of getting 2 heads and S₂ is the event of not getting 2 heads on tossing 6 coins

Now, Probability of statement to be truth if 2 heads are obtained on tossing of 6 coins

∵ Person speaks truth 3 times out of 5.
Similarly, Probability of statement to be not truth if 2 heads are obtained on tossing of 6 coins
Now, the probability that this event is actually true,

Question 27.
In throwing two dices, what is the probability that neither same digit appear nor sum of digit be 9 ?
Solution:
Total events in throwing two dices simultaneously
n(S) = 6² = 36
Total events to get same digit at two dices and sum 9
A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (4, 5), (5, 4), (6, 3)}
n(A) = 10
Probability to get same digit on dices or sum 9

Thus, required probability

Question 28.
Three coins are tossed simultaneously, then find the probability where as –
(i) Exactly two heads appear.
(ii) At least two heads appear,
(iii) Maximum two head appear.
(iv) AH the three are heads.
Solution:
In throwing three coins obtained sample space
S = {(HHH), (HHT), (7/77/7), (HTT), (THH), (THT), (TTH), (TTT)}
n(s) = 8
(i) Probability to get exactly two heads = 3/8
(ii) Events if at least two heads apear
A = {(HHH), (HHT), (HTH), (THH)}
∴ n(A) = 4
Thus, required probability

(iii) Event to get maximum 2 heads –
A = {(HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}
n(A) = 7
Thus, required probability

(iv) Events to get all the three heads
A = {HHH}
n(A) = 1
Thus, required probability

Question 29.
In a horse race, four horses A, B, C, D run odds in favour of A, B, C, D are respective of 1 : 3, 1 : 4, 1: 5 and 1: 6. Find the probability that are one of them wins.
Solution:
Let event to win horses A, B, C, D are E, F, G, H respectively.

Since only one is the winner
Thus, these are mutually exclusive events
∴ Required probability

Question 30.
Probability that a person will alive in next 25 years is 3/5 and of his wife 2/3. Find the following probability:
(i) Both remains alive.
(ii) None of two remains alive.
(iii) At least one remains alive.
(iv) Only wife remains alive.
Solution:
According to question, prob. of one person alive

(i) Prob. that both remains alive
By formula P(AB) = P(A).P(B)
P(EF) = P(E).P(F)
= 3/5 × 2/3 = 2/5
(ii) Probability that none of them remain alive

(iii) Probability that at least one remain alive

(iv) Prob. that only wife remain alive

Question 31.
A and B are independent witness. Probability that A speaks truth is x and y of B if A agree with B for any statement, then prove that probability of the truth of statement will be xy/(1 – x + 2xy).
Solution:
5 The prob. that A speaks truth
P(A) = x
Probability that B speaks truth
P(B) = y
If both are agree with any statement then, prob. that statement is true

Question 32.
Three males A,B C tossed a coin one by one. Who gets tail first he will win. If A has first chance, then what is the probability to winning of A.
Solution:
Prob. to get tail = 1/2
and Pro. to not get tail = 1/2
∵ A toss a coin in first chance thus he may win 1st time, 4th time, 7th time.
∴ Prob. that A win

Similarly prob. that B wins

Similarly prob. that C wins

Question 33.
Sulakshina and Sunayana tossed a coin one by one. Who get tail first she will win.If Sulakshina has first chance, then find probability that both of them win.
Solution:
Prob. to get.tail = 1/2
and prob. to not get tail = 1/2
∵ Since first, sulakshina toss a coin
Thus, she may win in 1st, 3rd, 5th attempt
∴ Probability that Sulakshina wins

and probability that sunayana wins

Question 34.
One digit is selected from the following two groups of numbers –
(1, 2, 3, 4, 5, 6, 7, 8, 9), (1, 2, 3, 4, 5, 6, 7, 8, 9)
If P₁ is sum of both digit as 10, P₂ is sum of both digit as 8, then find P₁ + P₂.
Solution:
P₁ = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}
n(P₁) = 9
Total ways = 9 × 9 = 81 = n(S)

Question 35.
If P(A) = 0.4, P(B) = 0.8, P(B/A) = 0.6, then find P(A/B) and P(A ∪ B).
Solution:
By formula

Question 36.
If P(E) = 0.35, P(P) = 0.45, P(E ∪F) = 0.65, then find P(F/E).
Solution:

Question 37.
A dice is thrown five times, find the probability getting only one.
Solution:
Probability to get 1 in throwing a dice = (1/6)
and Probability not get = 1 – 1/6 = 5/6
Thus probability to get only digit 1 in five tosses

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