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CBSE Class 12th Mathematics

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Chapter-13Probability

Important Questions

Q.1 A card is draw at random from deck of 52 cards. Find the probability its being a spadeor king.

Sol.

𝐹𝑜𝑟 𝑎 𝑑𝑒𝑐𝑘 𝑜𝑓 𝑐𝑎𝑟𝑑𝑠

𝑛(𝑆) = 52

𝐿𝑒𝑡 𝐴 = 𝐸𝑣𝑒𝑛𝑡 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑠𝑝𝑎𝑑𝑒 𝑐𝑎𝑟𝑑

𝐵 = 𝐸𝑣𝑒𝑛𝑡 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑘𝑖𝑛𝑔

∴ 𝐴 ∩ 𝐵) = 𝐸𝑣𝑒𝑛𝑡 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑘𝑖𝑛𝑔 𝑜𝑟 𝑠𝑝𝑎𝑑𝑒

𝑁𝑜𝑤,𝑛 (𝐴) = 13 (𝑠𝑝𝑎𝑑𝑒), 𝑛(𝐵) = 4(𝑘𝑖𝑛𝑔)

𝑎𝑛𝑑 𝑛(𝐴 ∩ 𝐵) = 1

∴ 𝑃(𝐴) =𝑛(𝐴)

𝑛(𝑆)=13

52=1

4

∴ 𝑃(𝐵) =𝑛(𝐵)

𝑛(𝑆)=4

52=1

13

𝑎𝑛𝑑 𝑃(𝐴 ∩ 𝐵) =𝑛(𝐴 ∩ 𝐵)

𝑛(𝑠)=1

52

𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦

𝑃 (𝑎 𝑠𝑝𝑎𝑑𝑒 𝑜𝑟 𝑘𝑖𝑛𝑔)

𝑝(𝐴 ∩ 𝐵) = 𝑝(𝐴) + 𝑃( 𝐵)− 𝑝(𝐴 ∩ 𝐵)

=1

4+1

13−1

52=4

13



Q.2 P speaks truth in 60% of the cases and Q in 70 % of the cases. In what percentage ofcase are they likely to contradict each other in stating the same fact?

Sol.

𝐿𝑒𝑡 𝐴 = 𝐸𝑣𝑒𝑛𝑡 𝑡ℎ𝑎𝑡 𝑃 𝑠𝑝𝑒𝑎𝑘𝑠 𝑡ℎ𝑒 𝑡𝑟𝑢𝑡ℎ

𝐵 = 𝐸𝑣𝑒𝑛𝑡 𝑡ℎ𝑎𝑡 𝑄 𝑠𝑝𝑒𝑎𝑘𝑠 𝑡ℎ𝑒 𝑡𝑟𝑢𝑡ℎ.

𝑎𝑛𝑑 �̅� = 𝑐𝑜𝑛𝑡𝑟𝑎𝑑𝑖𝑐𝑡𝑜𝑟𝑦 𝑒𝑣𝑒𝑛𝑡, 𝑖. 𝑒.𝑃 𝑠𝑝𝑒𝑎𝑘𝑠 𝑙𝑖𝑒

𝐵 = 𝑐𝑜𝑛𝑡𝑟𝑎𝑠𝑖𝑐𝑡𝑜𝑟𝑦 𝑒𝑣𝑒𝑛𝑡, 𝑖. 𝑒.𝑄 𝑠𝑝𝑒𝑎𝑘𝑠 𝑙𝑖𝑒

𝑁𝑜𝑤,𝑃(𝐴) =60

100=3

5

∴ 𝑃(�̅�) = 1− 𝑃(𝐴) = 1 −3

5=2

5

𝑎𝑛𝑑 𝑃(𝐵) =70

100=7

10

∴ 𝑃(𝐵�) = 1− 𝑃(𝐵) = 1 −7

10=3

10

𝑁𝑜𝑤, 𝑐𝑜𝑛𝑡𝑟𝑎𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑖𝑠𝑒𝑠

𝑊ℎ𝑒𝑛 (𝑖) 𝑃 𝑠𝑜𝑒𝑎𝑗𝑠 𝑡𝑔𝑒 𝑡𝑟𝑦𝑡𝑔 𝑎𝑏𝑑 𝑄 𝑡𝑒𝑙𝑙𝑠 𝑙𝑖𝑒

𝑖. 𝑒.𝐴 ∩ 𝐵�

(𝑖𝑖) 𝑃 𝑡𝑒𝑙𝑙𝑠 𝑙𝑖𝑒 𝑎𝑛𝑑 𝑄 𝑠𝑝𝑒𝑎𝑘𝑠 𝑡ℎ𝑒 𝑡𝑟𝑢𝑡ℎ

𝑖. 𝑒. �̅� ∩ 𝐵

𝐻𝑒𝑛𝑐𝑒, 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦

= 𝑃(𝐴 ∩ 𝐵�) + 𝑃(�̅� ∩ 𝐵)

= 𝑃(𝐴)𝑃(𝐵�) + 𝑃(�̅�)𝑃(𝐵)

=3

5×3

10+2

5×7

10=23

50



Q.3 There are three bags A, B and C. Bag A contains 4 white balls and 5 blue balls. Bag Bcontains 4 white balls and 3 blue balls. Bag C contains 2 white balls and 4 blue balls. Oneball is drawn form each of these bags. What is the probability that ine of these three ballsdrawn, two are white balls and one is a blue ball?

Sol.

𝐿𝑒𝑡,

𝑋 = 𝐸𝑣𝑒𝑛𝑡 𝑜𝑓 𝑑𝑟𝑎𝑤𝑖𝑛𝑔 𝑤ℎ𝑖𝑡𝑒 𝑏𝑎𝑙𝑙𝑠 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐴

𝑌 = 𝐸𝑣𝑒𝑛𝑡 𝑜𝑓 𝑑𝑟𝑎𝑤𝑖𝑛𝑔 𝑤ℎ𝑖𝑡𝑒 𝑏𝑎𝑙𝑙𝑠 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐵

𝑍 = 𝐸𝑣𝑒𝑛𝑡𝑠 𝑜𝑓 𝑑𝑟𝑎𝑤𝑖𝑛𝑔 𝑤ℎ𝑖𝑡𝑒 𝑏𝑎𝑙𝑙𝑠 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐶

𝑇ℎ𝑒𝑛,𝑃(𝑋) =4

9,𝑃(𝑌) =

4

7𝑎𝑛𝑑 𝑃(𝑍) =

2

3=1

3

∴ 𝑃( 𝑋�) = 𝐵𝑎𝑙𝑙 𝑑𝑟𝑎𝑤𝑛 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐴 𝑖𝑠 𝑏𝑙𝑢𝑒

= 1 − 𝑃(𝑋) = 1 −4

9=5

9

𝑃 (𝑌�) = 𝐵𝑎𝑙𝑙𝑠 𝑑𝑟𝑎𝑤𝑛 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐴 𝑖𝑠 𝑏𝑙𝑢𝑒

= 1 − 𝑃(𝑋) = 1 −4

7=3

7

𝑃(�̅�) = 1−1

3=2

3

𝑁𝑜𝑤, 𝑡𝑤𝑜 𝑤ℎ𝑖𝑡𝑒 𝑏𝑎𝑙𝑙𝑠 𝑎𝑛𝑑 𝑖𝑛𝑒 𝑏𝑙𝑢𝑒 𝑏𝑎𝑙𝑙𝑐𝑎𝑛𝑑 𝑏𝑒 𝑑𝑟𝑎𝑤𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑚𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝑒𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒 𝑤𝑎𝑦𝑠:

(𝑎) 𝑤ℎ𝑖𝑡𝑒 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐴,𝑤ℎ𝑖𝑡𝑒 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐵𝑎𝑛𝑑 𝑏𝑙𝑢𝑒 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐶, 𝑖. 𝑒.𝑋 ∩ 𝑌 ∩ 𝑍̅.


= 𝑃(𝑎) + 𝑃(𝑏) + 𝑃(𝑐)

(𝑏) 𝑤ℎ𝑖𝑡𝑒 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐴,𝑤ℎ𝑖𝑡𝑒 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐵 𝑎𝑛𝑑 𝑤ℎ𝑖𝑡𝑒 𝑓𝑟𝑜𝑚𝐶 𝑖. 𝑒,𝑋 ∩ 𝑌� ∩ 𝑍

(𝑐)𝑏𝑙𝑢𝑒 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐴,𝑤ℎ𝑖𝑡𝑒 𝑓𝑟𝑜𝑚 𝑏𝑎𝑔 𝐵 𝑎𝑛𝑑 𝑤ℎ𝑖𝑡𝑒 𝑓𝑟𝑜𝑚 𝐶 𝑖. 𝑒.𝑋� ∩ 𝑌 ∩ 𝑍




= 𝑃(𝑎) + 𝑃(𝑏) + 𝑃(𝑐)

= 𝑃(𝑋 ∩ 𝑌 ∩ �̅�) + 𝑃(𝑋 ∩ 𝑌� ∩ 𝑍) + 𝑃(𝑋� ∩ 𝑌 ∩ 𝑍)

= 𝑃(𝑋)𝑃(𝑌)𝑃(�̅�) + 𝑃(𝑋) 𝑃(𝑌�) 𝑃(𝑍) + 𝑃(𝑋�)𝑃(𝑌)𝑃(𝑍)

= �4

9×4

5×2

3+4

9×3

7×1

3+5

9×4

7×1

3� =

64

189

Q.4 Five disc are thrown simultaneously. If the occurrence of an even number in a singledisc is considered a success. Find the probability of almost 3 successes.

Sol.

𝐿𝑒𝑡 𝑋 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠 𝑖𝑛 5 𝑡ℎ𝑟𝑜𝑤𝑠 𝑜𝑓 𝑎 𝑑𝑖𝑒.

𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟

𝑖 𝑒,𝑃 =3

6=1

2

𝑁𝑜𝑤,𝑋 𝑖𝑠 𝑎 𝑏𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑡𝑒 𝑤𝑖𝑡ℎ 𝑟 𝑠𝑢𝑐𝑐𝑒𝑠𝑠

𝐻𝑒𝑟𝑒,𝑛 = 5,𝑝 =1

2, 𝑞 =

1

2

∴ 𝑃(𝑋 = 𝑟) = 5rC �1

2�r

�1

2�5−r

𝑊ℎ𝑒𝑟𝑒, 𝑟 = 0,1,2,… . .5

= 5rC �1

2�5

∴ 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 𝑃(𝑋 ≤ 3) = 1 − 𝑝(𝑥 ≥ 3)

= 1 − {𝑝(𝑋 = 4)+ 𝑃(𝑥 = 5)}

5 55 5 5

4 5

1 11

2 2rC C C� ��

� � � ��



1 − �5

32+1

32� =

26

32=13

16

Q.5 Find the probability distribution of the number of head when three coins are tossed.

Sol.

𝐿𝑒𝑡 𝑝 = 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 ℎ𝑒𝑎𝑑 =1

2

𝑎𝑛𝑑 𝑞 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑛𝑜 ℎ𝑒𝑎𝑑 = 1 −1

2=1

2

𝑆𝑢𝑝𝑝𝑜𝑠𝑒,𝑋 𝑑𝑒𝑛𝑜𝑡𝑒 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 ℎ𝑒𝑎𝑑𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑖𝑛 𝑡ℎ𝑟𝑒𝑒 𝑡𝑜𝑠𝑠𝑒𝑠 𝑜𝑓 𝑎 𝑐𝑜𝑖𝑛.

𝐻𝑒𝑟𝑒,𝑛 = 3,𝑝 =1

2, 𝑞 =

1

2

� �3

3

33

1 12 2

Where, 0,1,2,3

12

r r

r

r

p X r C

r

C

��

�

� ��

𝑇ℎ𝑢𝑠, 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑋 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠

𝑋 0 1 2 3

𝑃(𝑋)3

30

12

C � ��

33

1

12

C � ��

33

2

12

C � ��

33

3

12

C � ��

𝐻𝑒𝑛𝑐𝑒,

𝑋 0 1 2 3

𝑃(𝑋) 1

8

3

8

3

8

1

8



Q.6 Find the binominal distribution for which the mean is 4 and variance is 3.

Sol.

𝑊𝑒 ℎ𝑎𝑣𝑒,

𝑀𝑒𝑎𝑛 𝑛𝑝 = 4 𝑎𝑛𝑑 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 3

𝑖. 𝑒. ,𝑛𝑝𝑞 = 3

∴ 𝑛𝑝𝑞 =3

4

𝑜𝑟 𝑞 =3

4∴ 𝑝 = 1 −

3

4=1

4

𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝 =1

4𝑖𝑛 𝑛𝑝 = 4

𝑛 ×1

4= 4 ⇒ 𝑛 = 4 × 4 = 16

𝑜𝑟 𝑛 = 16

𝑆𝑜, 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑏𝑖𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦

� �

� � � ��

1616

n

1 34 4

Where, 0,1,2,.......16

(Using p)

r r

r

n rr r

p X r C

r

C p q

�

�

� � � ��

�

Q.7 The sum and the product of the mean and variance of a binomial distribution are 24 and 128 respectively. Find the distribution.

Sol.

𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡

𝑚𝑒𝑎𝑛 = 𝑛𝑝 𝑎𝑛𝑑 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑛𝑝𝑞 𝑎𝑛𝑑 𝑞 = 1 − 𝑃

𝐴𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛



𝑚𝑒𝑎𝑛 + 𝑣𝑎𝑟𝑖𝑒𝑛𝑐𝑒 = 24

𝑛𝑝+ 𝑛𝑝𝑞 = 24

𝑜𝑟 𝑛𝑝 (1 + 𝑞) = 24 ⇒ 𝑛𝑝 =24

1 + 𝑞. . . (1)

𝑎𝑛𝑑 𝑚𝑒𝑎𝑛 𝑥 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 128

𝑜𝑟 𝑛𝑝 × 𝑛𝑝𝑞 = 128

𝑜𝑟 𝑛2 𝑝2 × 𝑞 = 128

𝑜𝑟 𝑛2 𝑝2 =128

𝑞… (2)

𝐹𝑟𝑜𝑚 (1)𝑎𝑛𝑑 (2),

�24

1 + 𝑞�2

=128

𝑞⇒

156

(1 + 𝑞)2=128

𝑞

𝑜𝑟 576𝑞 = 128(𝐼 + 𝑞)2

𝑜𝑟 9𝑞 = 2( 1 + 2𝑞 + 𝑞2)

𝑜𝑟 9𝑞 = 2 + 4𝑞 + 2𝑞 + 𝑞2) 𝑜𝑟 2𝑞2 + 4𝑞 + 2 − 9𝑞 = 0

𝑝𝑟 2𝑞2− 5𝑞 + 2 = 0 𝑜𝑟 (2𝑞 − 1)(𝑞 − 2) =

𝑜𝑟 𝑞 =1

2𝑞 = 2 {𝐵𝑢𝑡 𝑞 ≠ 2}

∴ 𝑞 =1

2

: .𝑝 = 1 −1

2=1

2

𝑁𝑜 𝑞,𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑛𝑝+ 𝑛𝑝𝑞 = 24

𝑜𝑟𝑛

2+𝑛

2= 24 ⇒ 𝑛 = 32 �𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑝 = 𝑞 =

1

2�

𝐻𝑒𝑛𝑐𝑒, 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑏𝑖𝑜𝑛𝑚𝑖𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑖𝑠



� �32

322

1 12 2

Where, 0,1,2,.......32

r r

p X r C

r

��

�

Q.8 If two dice are rolled 12 times, obtain the mean and the variance of the distribution ofsuccesses, if getting a total greater than 4 is considered a success.

Sol.

𝐿𝑒𝑡 𝑝

= 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑡𝑜𝑡𝑎𝑙 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 4 𝑖𝑛 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑡ℎ𝑟𝑜𝑤 𝑜𝑓 𝑎 𝑝𝑎𝑖𝑟 𝑜𝑓 𝑑𝑖𝑐𝑒.

1 − 𝑝 (𝑡𝑜𝑡𝑎𝑙 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 4) = 1 −6

36=5

6

∴ 𝑞 = 1 −5

6=1

6

∴ 𝑚𝑒𝑎𝑛 = 𝑚𝑝 =5

6× 12 = 10

𝑎𝑛𝑑 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑛𝑝𝑞 = 12 ×5

6×1

6=5

3

Q.9 A company has two plant to manufacture scooters. Plant I manufacture 70% of thescooters and plant II manufactures 30%. At Plant I, 80% of the scooters are rated as of thestandard quality and at plant II 90% of the scooters are rated as of standard quality. A scooter is choosen at random and is found to be of be of standard quality. What is theprobability that it has come from plant II.

Sol.

𝐿𝑒𝑡 𝐸1,𝐸2 𝑎𝑛𝑑 𝐴 𝑏𝑒 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑒𝑣𝑒𝑛𝑡𝑠:

𝐸1 = 𝑝𝑙𝑎𝑛𝑡 𝐼 𝑖𝑠 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑

𝐸2 = 𝑝𝑙𝑎𝑛𝑡 𝐼𝐼 𝑖𝑠 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑

𝑎𝑛𝑑 𝐴 = 𝑠𝑐𝑜𝑜𝑡𝑒𝑟 𝑜𝑓 − 𝑠𝑡𝑎𝑛𝑑𝑒𝑟𝑑 𝑞𝑢𝑎𝑙𝑖𝑡𝑦.



𝐻𝑒𝑟𝑒,𝑃(𝐸1) =70

100

𝑃(𝐸2) =30

100,𝑃(𝐴/𝐸1) =

80

100

𝑎𝑛𝑑 𝑃 (𝐴/𝐸2) =90

100

𝑁𝑜𝑤,𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑠𝑐𝑜𝑜𝑡𝑒𝑟 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑓𝑟𝑜𝑚 𝑝𝑙𝑎𝑛𝑡 𝐼1, 𝑖. 𝑒,𝑃(𝐸2/𝐴) =?

𝑁𝑜𝑤,𝑢𝑠𝑖𝑛𝑔 𝐵𝑎𝑦𝑒′𝑠 𝑡ℎ𝑟𝑜𝑟𝑒𝑚

𝑃(𝐸2/𝐴) =𝑃(𝐸2)𝑃(𝐴/𝐸2)

𝑃(𝐸1)𝑃(𝐴/𝐸1) + 𝑃(𝐸2)𝑃(𝐴/𝐸2)

=

30100

×90100

70100

×80100

×30100

×90100

=27

56 + 27=27

83

Q.10 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truckdrivers. The probability of an accident involving a scooter driver, car driver and a truckdriver is 0.01, 0.03 and 0.15 respectively. One of the insured person meets with an accident.what is the probability that he is a scooter driver?

Sol.

𝐿𝑒𝑡 𝐸1,𝐸2,𝐸3𝑎𝑛𝑑 𝐴 𝑏𝑒 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑒𝑣𝑒𝑛𝑡𝑠:

𝐸1 = 𝑃𝑒𝑟𝑠𝑜𝑛 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑖𝑠 𝑎 𝑠𝑐𝑜𝑜𝑡𝑒𝑟 𝑑𝑟𝑖𝑣𝑒𝑟

𝐸2 = 𝑃𝑒𝑟𝑠𝑜𝑛 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑖𝑠 𝑎 𝑐𝑎𝑟 𝑑𝑟𝑖𝑣𝑒𝑟

𝐸3 = 𝑃𝑒𝑟𝑠𝑜𝑛 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑖𝑠 𝑎 𝑡𝑟𝑢𝑐𝑘 𝑑𝑟𝑖𝑣𝑒𝑟

𝐴 = 𝑃𝑒𝑟𝑠𝑜𝑛 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑𝑤𝑖𝑡ℎ 𝑎𝑛 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡

𝑇𝑜𝑡𝑎𝑙 𝑝𝑒𝑟𝑠𝑜𝑛𝑠 = 2000+ 4000 + 6000 = 12000



𝑃(𝐸1) =2000

12,000=1

6

𝑃(𝐸2) =4000

12,000=1

3

𝑃(𝐸3) =6000

12,000=1

2

𝑁𝑜𝑤,𝑃(𝐴/𝐸1) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑎 𝑝𝑒𝑟𝑠𝑜𝑛 𝑚𝑒𝑒𝑡𝑠 𝑤𝑖𝑡ℎ 𝑎𝑛 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡 𝑖𝑠 𝑎 𝑠𝑐𝑜𝑜𝑡𝑒𝑟 𝑑𝑟𝑖𝑣𝑒𝑟

= 0.01

𝑃(𝐴/𝐸2) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 ~𝑛 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡 𝑜𝑓 𝑐𝑎𝑟 𝑑𝑟𝑖𝑣𝑒𝑟 = 0.03

𝑎𝑛𝑑 𝑃(𝐴/𝐸3) = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎𝑛 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡 𝑜𝑓 𝑎 𝑡𝑟𝑢𝑐𝑘 𝑑𝑟𝑖𝑣𝑒𝑟 = 0.15

𝑁𝑜𝑤,𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑝𝑒𝑟𝑠𝑜𝑛 𝑖𝑠 𝑎 𝑠𝑐𝑜𝑜𝑡𝑒𝑟 𝑑𝑟𝑖𝑣𝑒𝑟

𝑖. 𝑒.𝑃 (𝐸1/𝐴) =?

𝑁𝑜𝑤,𝑢𝑠𝑖𝑛𝑔 𝐵𝑎𝑦𝑒′𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚

𝑃(𝐸1/𝐴)′

=𝑃(𝐸2)𝑃(𝐴/𝐸1)

𝑃(𝐸1)𝑃(𝐴/𝐸1)+ 𝑃(𝐸2)𝑃(𝐴/𝐸2)+ 𝑃(𝐸3)𝑃(𝐴/𝐸3)

=

16 × 0.01

16 × 0.01 +

13 × 0.03 +

12 × 0.15

1

1 + 6 + 45=1

52

Q.11 In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour.Find E(X) and Var (X).

Sol.



P(X = 0) = 30% = 0.3

P(X = 1) = 70% = 0.7

Now,

E (X) = � �i iX P X�0 0.3 1 0.70.7

� � � ��

E (X2) = � �2i iX P X�

2 20 0.3 1 0.70.7

� � � ��

� � � � � �

� �

22

2 0.7 0.7

0.21

Var X E X E X� � � ��

� �

�

Q.12 A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16,19 and 20 years. One student is selected in such a manner that each has the same chance ofbeing chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

Sol.

Total students = 15

Probability of each student to be selected =1

15

P(X = 14) =2

15, P(X = 15) =

115

, P(X = 16) =2

15, P(X = 17) =

315

,



P(X = 18) =1

15, P(X = 19) =

215

, P(X = 20) =3

15, P(X = 21) =

115

Mean of X = E (X) = � �i iX P X�2 1 2 3 1 2 3 1

14 15 16 17 18 19 20 2115 15 15 15 15 15 15 15

26315

17.5

� � � � � � � � � � � � � � � �

�

�

E (X2) = � �2i iX P X�

2 2 2 2 2 2 2 22 1 2 3 1 2 3 114 15 16 17 18 19 20 21

15 15 15 15 15 15 15 15468315

� � � � � � � � � � � � � � � �

�

312.2�

� � � � � �

� �

22

2 312.2 17.5

4.78

Var X E X E X� � � ��

� �

�

� �Standard deviation 4.78 2.18Var X� � �

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