PROBABILITY QUESTIONS FOR GRADE 11

Question 1 :

What is the chance that (i) non-leap year (ii) leap year should have fifty three Sundays?

Solution :

Number of days in non leap year  =  365

  =  52 week + 1 day

  =  52 Sundays + 1 day

Number of Sundays in a year  =  52

Days in a week  =  { Sunday, Monday, Tuesday, Wednesday, Thursday, Saturday }

n(S)  =  7

(i)  Let "A" be the event of getting the probability of 53 Sundays in a year.

n(A)  =  1

P(A)  =  n(A) / n(S)

  =  1/7

(ii) Let "B" be the event of getting the probability of 53 Sundays in a leap year.

Number of days in leap year  =  366 days

  =  364 days + 2 days

  =  52 Sundays + 2 days

Chance of being 53 Sundays  =  {(Sun, mon),(mon, tue), (tue, wed), (wed, thur) (thur, fri) (fri, sat) (sat, sun)}

B  =  {(Sun, mon), (sat, sun)}

n(B)  =  2

P(B)  =  n(B) / n(S)

P(B)  =  2/7

Question 2 :

Eight coins are tossed once, find the probability of getting (i) exactly two tails (ii) at least two tails (iii) at most two tails

Solution :

When eight coins are tossed, number of events will be  =  2⁸  =  256

n(S)  =  256

(i) Let "A" be the event of getting two tails.

n(A)  =  ⁸c₂

  n(A)  =  (8 ⋅ 7)/(2 ⋅ 1) 

n(A)  =  28

P(A)  =  n(A) / n(S)

  =  28 / 256

  =  7/64

(ii) at least two tails

Let "B" be the event of getting at least two tails.

At least two tails means, we may get 2 tails, 3 tails  so on up to 8 tails.

Other wise,

  =  1 - P(Getting lesser than two tails)

  =  1 - [P(0 tail) + P(1 tail)]

  =  1 - [(⁸c₀/256) + (⁸c₁/256)]

  =  1 - [(1/256) + (8/256)]

  =  1 - (9/256)

  =  (256 - 9) / 256

  =  247 / 256

(iii) at most two tails

Let "C" be the event of getting at most two tails.

At most two tails means, we may get 0 tail, 1 tail and 2 tails.

  =   P(0 tail) + P(1 tail) + P(1 tails)

  =  (⁸c₀/256) + (⁸c₁/256) + (⁸c₂/256)

  =  (1/256) + (8/256) + (28/256)

  =  37/256

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