The probability I call my friend Katie on a given day is 03

The probability I call my friend Katie on a given day is 0.3, and the probability I call my friend Rachel on a given day is 0.3. Fur- thermore, the probability that I call either Rachel or Katie on a given day is 0.9.

(i). Are the events of ‘calling Katie’ and ‘calling Rachel’ indepen- dent? Justify your answer.

(ii). Assume that I do not call Katie or Rachel multiple times on a given day. Consider a sequence of five days, and let X denote the number of times I call either Katie or Rachel in those five days. What kind of random variable is X? Characterize the random variable X.

(iii). Calculate the probability that I call Katie or Rachel exactly three times in five days.

(iv). Calculate the probability that I call Katie or Rachel at least three times in five days?

Solution

Let us consider A be the event that  I call my friend Katie on a given day.

P(A) = 0.3

and B be the event that I call my friend Rachel on a given day.

P(B) = 0.3

   the probability that I call either Rachel or Katie on a given day is 0.9.

that is P(A or B) = 0.9

P(A or B) = P(A) + P(B) - P(A and B)

0.9 = 0.3 + 0.3 - P(A and B)

0.9 = 0.6 - P(A and B)

P(A and B) = 0.3

Independent evevnts :

P(A and B) = P(A) * P(B)

P(A)*P(B) = 0.3 * 0.3 = 0.09

P(A and B) = 0.3

P(A and B) P(A) * P(B)

Events A and B are not independent.

Assume that I do not call Katie or Rachel multiple times on a given day.

Consider a sequence of five days, and let X denote the number of times I call either Katie or Rachel in those five days.

X isa discrete binomial random variable with n=5 and p = 0.9

Calculate the probability that I call Katie or Rachel exactly three times in five days.

that is here we have to calculate P(X = 3)

The probability mass function of X is,

P(X=x) = (nCx) * p^x * q^n-x   ( C is combination)

n=5

p=0.9

q=1-p = 1 - 0.9 = 0.1

P(X=3) = (5C3) * 0.9³ * 0.1² = 0.0729

Calculate the probability that I call Katie or Rachel at least three times in five days?

P(X >=3) = 1 - P(X < 3)

= 1 - [P(X=0) + P(X=1) + P(X=2) ]

P(X=0) = (5C0) * 0.9⁰ * 0.1⁵ = 0.00001

P(X=1) = (5C1) * 0.9¹ * 0.1⁴ = 0.00045

P(X=2) = (5C2) * 0.9² * 0.1³ = 0.0081

P(X >=3) = 1 - [ 0.00001 + 0.00045 + 0.0081 ]

= 1 - 0.00856

=0.99144