# CONDITIONAL PROBABILITY WORD PROBLEM EXAMPLES

Example 1 :

A year is selected at random. What is the probability that (i) it contains 53 Sundays (ii) it is a leap year which contains 53 Sundays

Solution :

In a ordinary year, we have 365 days

=  52 weeks + 1 day

=  52 Sundays + 1 day

That one day could be = {Sunday, Monday, Tuesday, Wednesday, Thursday, Saturday}

Probability of getting 53 Sundays in ordinary year  =  1/7

In a leap year, we have 366 days

=  52 weeks + 2 days

=  52 Sundays + 2 days

That one day could be = {(Sun, mon) (mon, tue) (tue, wed)(wed, thu) (thu, fri) (fri, sat) (sat, sun)}

Probability of getting 53 Sundays in leap year  =  2/7

(i) it contains 53 Sundays

Hint in the given question : A year is selected at random

From this, we come to know that we may choose either ordinary year or leap year.

P(getting 53 Sundays)

= P(selecting ordinary year) ⋅ P(53 Sundays in ordinary year)

=  (3/4) ⋅ (1/7)

[(A year occurring  once in 4 years is know as leap year)

So, probability of being ordinary year  =  3/4]

=  3/28  -------(1)

= P(selecting leap year) ⋅ P(53 Sundays in leap year)

=  (1/4) ⋅ (2/7)

=  2/28  -------(2)

(1) + (2)

= (3/28) + (2/28)

=  5/28

(ii) it is a leap year which contains 53 Sundays

=  P(selecting leap year) ⋅ P(53 Sundays in leap year)

=  (1/4) ⋅ (2/7)

=  1/14

Example 2 :

Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?

Solution :

P(Hitting a target by X)  =  3/4, P(X')  =  1/4

P(Hitting a target by Y)  =  4/5, P(Y')  =  1/5

P(Hitting a target by Z)  =  2/3, P(Z')  =  1/3

P(target is damaged by exactly 2 hits)

=  P(X'nYnZ) + P(XnY'nZ) + P(XnYnZ')

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

=  (1/4)(4/5)(2/3) + (3/4)(1/5)(2/3) + (3/4)(4/5)(1/3)

=  8/60 + 6/60 + 12/60

=  26/60

=  13/30

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