Question

seeta and geeta are two sisters with an age difference of 3 years find the probability that they are born in a leap year and at least one of them is born in a leap year

Answer 1

Answer:

Probability of bith born in leap year  = 0

Probability of atleat one is born in leap year = 1/2

Step-by-step explanation:

Leap year comes after every 4 years. As the difference between two sisters is 3 years, even if one sister is born on leap year, other cannot be born on leap year.

Thus - Probability of both the sisters born on leap year = 0

As the leap year comes every 4 years, indipendent probability for one sister to be born on leap year P(s1) = 1/4

For second sister independent probaibility to be born on leap year

P(s2) = 1/4

By the laws of probability, either one sister to be born on leap year is the sum of the two probabilities.

P(s) = P(s1) + P(s2)

P(s) = 1/4 + 1/4  = 1/2

P(s) = 1/2

Probability that atleast one of the sister is born in a leap year = 1/2

Answer 2

Answer: (i) 0

(ii) 1/2

Step-by-step explanation:

(I) As the age difference is 3 years between both the sisters and a leap year occurs every 4 year ...... so , it is an impossible event that both are born in a leap year.........

P(E) = 0

(II) now let us assume that the elder sister is born in the year x.

So , the smaller one is born in the year x+3.

Total possible outcomes are :

(I) x is a leap year and x+4 is also a leap year.

(II) x+1 is a leap year and x+5 is a lena year.

(III) x+2 is a leap year and x+6 is a lena year.

(IV) x+3 is a leap year and x + 7 is a leap year.

Now , out of the above four , only (I) and (IV) are such outcomes where at least a sister is born in a leap year.

So , favourable outcomes are 2

Hence , P(E) = 2/4

= 1/2