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RD Chapter 33- Probability Ex-33.3 Interview Questions Answers

Question 1 : Which of the following cannot be validassignment of probability for elementary events or outcomes of sample space S ={w1, w2, w3, w4, w5, w6,w7}:

Answer 1 :

For each event to be avalid assignment of probability.

The probability ofeach event in sample space should be less than 1 and the sum of probability ofall the events should be exactly equal to 1.

(i) It is valid aseach P (wi) (for i=1 to 7) lies between 0 to 1 and sum of P (w1)=1

(ii) It is valid aseach P (wi) (for i=1 to 7) lies between 0 to 1 and sum of P (w1)=1

(iii) It is not validas sum of P (wi) = 2.8 which is greater than 1

(iv) it is not validas P (w7) = 15/14 which is greater than 1

Question 2 :
A die is thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3

Answer 2 :

Given: A die is thrown.
The total number of outcomes is six, n (S) = 6
By using the formula,
P (E) = favourable outcomes / total possible outcomes
(i) Let E be the event of getting a prime number
E = {2, 3, 5}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 6
= ½
(ii) Let E be the event of getting 2 or 4
E = {2, 4}
n (E) = 2
P (E) = n (E) / n (S)
= 2 / 6
= 1/3
(iii) Let E be the event of getting a multiple of 2 or 3
E = {2, 3, 4, 6}
n (E) = 4
P (E) = n (E) / n (S)
= 4 / 6
= 2/3

Question 3 :
In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) an even number on first
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) neither a doublet nor a total of 10
(xiii) odd number on the first and 6 on the second
(xiv) a number greater than 4 on each die
(xv) a total of 9 or 11
(xvi) a total greater than 8

Answer 3 :

Given: a pair of dice has been thrown, so the number of elementary events in sample space is 62 = 36
n (S) = 36
By using the formula,
P (E) = favourable outcomes / total possible outcomes
(i) Let E be the event that the sum 8 appears
E = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n (E) = 5
P (E) = n (E) / n (S)
= 5 / 36
(ii) Let E be the event of getting a doublet
E = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(iii) Let E be the event of getting a doublet of prime numbers
E = {((2, 2) (3, 3) (5, 5)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(iv) Let E be the event of getting a doublet of odd numbers
E = {(1, 1) (3, 3) (5, 5)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(v) Let E be the event of getting sum greater than 9
E = {(4,6) (5,5) (5,6) (6,4) (6,5) (6,6)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(vi) Let E be the event of getting even on first die
E = {(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
n (E) = 18
P (E) = n (E) / n (S)
= 18 / 36
= ½
(vii) Let E be the event of getting even on one and multiple of three on other
E = {(2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4) (3,6) (6,2) (6,4)}
n (E) = 11
P (E) = n (E) / n (S)
= 11 / 36
(viii) Let E be the event of getting neither 9 or 11 as the sum
E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(ix) Let E be the event of getting sum less than 6
E = {(1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3) (3,1) (3,2) (4,1)}
n (E) = 10
P (E) = n (E) / n (S)
= 10 / 36
= 5/18
(x) Let E be the event of getting sum less than 7
E = {(1,1) (1,2) (1,3) (1,4) (1,5) (2,1) (2,2) (2,3) (2,4) (3,1) (3,2) (3,3) (4,1) (4,2) (5,1)}
n (E) = 15
P (E) = n (E) / n (S)
= 15 / 36
= 5/12
(xi) Let E be the event of getting more than 7
E = {(2,6) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4) (5,5) (5,6) (6,2) (6,3) (6,4) (6,5) (6,6)}
n (E) = 15
P (E) = n (E) / n (S)
= 15 / 36
= 5/12
(xii) Let E be the event of getting neither a doublet nor a total of 10
E′ be the event that either a double or a sum of ten appears
E′ = {(1,1) (2,2) (3,3) (4,6) (5,5) (6,4) (6,6) (4,4)}
n (E′) = 8
P (E′) = n (E′) / n (S)
= 8 / 36
= 2/9
So, P (E) = 1 – P (E′)
= 1 – 2/9
= 7/9
(xiii) Let E be the event of getting odd number on first and 6 on second
E = {(1,6) (5,6) (3,6)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(xiv) Let E be the event of getting greater than 4 on each die
E = {(5,5) (5,6) (6,5) (6,6)}
n (E) = 4
P (E) = n (E) / n (S)
= 4 / 36
= 1/9
(xv) Let E be the event of getting total of 9 or 11
E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(xvi) Let E be the event of getting total greater than 8
E = {(3,6) (4,5) (4,6) (5,4) (5,5) (5,6) (6,3) (6,4) (6,5) (6,6)}
n (E) = 10
P (E) = n (E) / n (S)
= 10 / 36
= 5/18

Question 4 : In a single throw of three dice, find the probability of getting a total of 17 or 18

Answer 4 :

Given: The dices arethrown.

By using the formula,

P (E) = favourableoutcomes / total possible outcomes

Total number of possibleoutcomes is 63=216

So, n (S) = 216

Let E be the event ofgetting total of 17 or 18

E = {(6, 6, 5) (6, 5,6) (5, 6, 6) (6, 6, 6)}

n (E) = 4

P (E) = n (E) / n (S)

= 4 / 216

= 1/54

Question 5 :
Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail

Answer 5 :

Given: Three coins aretossed together.

By using the formula,

P (E) = favourableoutcomes / total possible outcomes

Total number ofpossible outcomes is 2= 8

(i) Let E be theevent of getting exactly two heads

E = {(H, H, T) (H, T,H) (T, H, H)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 8

(ii) Let E be theevent of getting at least two heads

E= {(H, H, T) (H, T,H) (T, H, H) (H, H, H)}

n (E)=4

P (E) = n (E) / n (S)

= 4 / 8

= ½

(iii) Let E be theevent of getting at least one head and one tail

E = {(H, T, T) (T, H,T) (T, T, H) (H, H, T) (H, T, H) (T, H, H)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 8

= ¾

Question 6 : What is the probability that an ordinary year has 53 Sundays?

Answer 6 :

Given: A year which includes 52 weeks and one day.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
So, we now have to determine the probability of that one day being Sunday
Total number of possible outcomes is 7
n(S) = 7
E = {M, T, W, T, F, S, SU}
n (E) = 1
P (E) = n (E) / n (S)
= 1 / 7

Question 7 : What is the probability that a leap year has 53 Sundays and 53 Mondays?

Answer 7 :

Given: A leap year which includes 52 weeks and two days
By using the formula,
P (E) = favourable outcomes / total possible outcomes
So, we now have to determine the probability of that remaining two days is Sunday and Monday
S = {MT, TW, WT, TF, FS, SSu, SuM}
n (S) = 7
E= {SuM}
n (E) = 1
P (E) = n (E) / n (S)
= 1 / 7

Question 8 :
A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:
(i) All the three balls are white
(ii) All the three balls are red
(iii) One ball is red and two balls are white

Answer 8 :

Given: A bag contains8 red and 5 white balls.

By using the formula,

P (E) = favourableoutcomes / total possible outcomes

Total number of waysof drawing three balls at random is 13C3

n (S) = 286

(i) Let E be theevent of getting all white balls

E= {(W) (W) (W)}

n (E)= 5C3=10

P (E) = n (E) / n (S)

= 10 / 286

= 5/143

(ii) Let E be theevent of getting all red balls

E = {(R) (R) (R)}

n (E)= 8C=56

P (E) = n (E) / n (S)

= 56 / 286

= 28/143

(iii) Let E be theevent of getting one red and two white balls

E = {(R…. 80th R)}

n (E)= 8C15C=80

P (E) = n (E) / n (S)

= 80 / 286

= 40/143

Question 9 : In a single throw of three dice, find the probability of getting the same number on all the three dice

Answer 9 :

Given: Three dice arerolled over.

By using the formula,

P (E) = favourableoutcomes / total possible outcomes

So, we now have todetermine the probability of getting the same number on all the three dice

Total number ofpossible outcomes is 63=216

n (S) = 216

Let E be the event ofgetting same number on all the three dice

E = {(1,1,1) (2,2,2)(3,3,3) (4,4,4) (5,5,5) (6,6,6)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 216

= 1/36

Question 10 : Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10

Answer 10 :

Given: Two unbiaseddice are thrown.

By using the formula,

P (E) = favourableoutcomes / total possible outcomes

So, we now have todetermine the probability of getting the sum of digits on dice greater than 10

Total number ofpossible outcomes is 62=36

n (S) = 36

Let E be the event ofgetting same number on all the three dice

E = {(5,6) (6,5)(6,6)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12


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RD Chapter 33- Probability Ex-33.3 Contributors

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