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

From given, we have,

In each match, assume that the Indian team has a 70% chance of winning.

Further, assuming that the matches are independent of each other.

Binomial distribution formula for calculating the probability is given as follows:

$b(x;n,P)=^nC_x \times P^x \times (1-P)^{n-x}$

where:  b = binomial probability (what we are trying to find)

x = total number of successes (wins for the Indian team)

P = probability of a success on an individual trial (70%)

n = number of games (5)

P(Win) is given as follows:

P(0)  $=^5C_0 \times (0.70)^0 \times (1-0.70)^{5-0}$ = (0.30)⁵ = 0.00243

P(1) $=^5C_1 \times (0.70)^1 \times (1-0.70)^{5-1}$= 5 ×  (0.70)¹ × (0.30)⁴ = 0.02835

P(2) $=^5C_2 \times (0.70)^2 \times (1-0.70)^{5-2}$= 10 × (0.70)² × (0.30)³ = 0.1323

P(3) $=^5C_3 \times (0.70)^3 \times (1-0.70)^{5-3}$= 10 × (0.70)³ × (0.30)² = 0.3087

P(4) $=^5C_4 \times (0.70)^4 \times (1-0.70)^{5-4}$= 5 × (0.70)⁴ × (0.30)¹ = 0.36015

P(5) $=^5C_5 \times (0.70)^5 \times (1-0.70)^{5-5}$= (0.70)⁵ = 0.16807

Therefore, we have:  

a. The Indian team will win the series?

P(Indian team wins the series) = P(Wins = 3,4, or 5)

=P(Wins=3)+P(Wins=4)+P(Wins=5) = 0.83692 = 83.692%

b. The team will win all five matches, and that the team will lose all?

P(Indian team wins all matches) = P(Wins = 5) = 16.807%

P(Indian team losses all matches) = P(Wins = 0) = 0.243%