Question

The Indian cricket team is visiting New Zealand to play a test series comprising five
matches. In each match, assume that the Indian team has a 70% chance of winning.
Further, assuming that the matches are independent of each other, what is the
probability that:
a. The Indian team will win the series?
b. The team will win all five matches, and that the team will lose all?

Answer 1

Answer:

THEY WILL WIN THE SERIES

THEY WILL NOT WIN ALL MATCHES THEY WILL LOSE ONE OR TWO MATCH

Answer 2

Answer:

a) P(X>=3)=0.8392

b) P(X=5)=0.16807

   P(X=0)=0.00243

Step-by-step explanation:

In each match, the probability of the Indian team winning is 0.7 and of the five matches each is independent of the other. Let X be a random variable representing the number of wins.

The probability of a win (success), p, is 0.7

The probability of a loss (failure), 1-p, is 0.3

The number of matches ( independent trials), n, is 5

Then the number of wins forms a Binomial distribution with p=0.7 and n=5

$X ~ Binomial(5,0.7)$

The probability function is;

$f(x)=P(X=x)= (nCx)*p^x*(1-p)^{(n-x)}, x=0,1,2,... \\\\f(x)=P(X=x)= (5Cx)*0.7^x*0.3^{(5-x)},x=0,1,2,3,4,5$

a) for the Indian team to win the series, the have to win at least 3 matches

$P(X\geq 3)=P(X=3)+P(X=4)+P(X=5)\\P(X\geq 3)= (5C3)*0.7^3*0.3^{(5-3)}+(5C4)*0.7^4*0.3^{(5-4)}+(5C5)*0.7^5*0.3^{(5-5)}\\P(X\geq 3)=0.83692$

b) Probability that the Indian team wins all 5 matches is

$P(X=5)=(5C5)*0.7^5*0.3^{(5-5)}=0.16807$

probability of the Indian team losing all 5 matches is;

$P(X=0)=(5C0)*0.7^0*0.3^{(5-0)}=0.00243$