RD Chapter 33- Probability Ex-33.1 |
RD Chapter 33- Probability Ex-33.2 |
RD Chapter 33- Probability Ex-33.4 |

**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 (w_{i}) (for i=1 to 7) lies between 0 to 1 and sum of P (w_{1})=1

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

(iii) It is not validas sum of P (w_{i}) = 2.8 which is greater than 1

(iv) it is not validas P (w_{7}) = 15/14 which is greater than 1

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

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

**Answer
4** :

Given: The dices arethrown.

By using the formula,

P (E) = favourableoutcomes / total possible outcomes

Total number of possibleoutcomes is 6^{3}=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

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^{3 }= 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

**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

**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

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 ^{13}C_{3}

n (S) = 286

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

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

n (E)= ^{5}C_{3}=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)= ^{8}C_{3 }=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…. 80^{th} R)}

n (E)= ^{8}C_{1}^{5}C_{2 }=80

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

= 80 / 286

= 40/143

**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 6^{3}=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

**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 6^{2}=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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