Question

3. Lok Sabha in India is to discuss a key constitutional amendment, and there will be four
rounds of voting. In each round of voting, assume that the ruling party has a 60% chance
of winning, i.e. getting the resolution passed by majority vote. Further assuming that the
voting rounds are independent of each other, what is the probability that:

a. The ruling party will win 0 rounds, I round, 2 rounds, 3 rounds or all 4 rounds of
voting?

b. The ruling party will win at least 1 round?
​

Answer 1

Answer:

If p = 0.6, q = 1 - 0.6 = 0.4,

Pn(k) = Cn(k)*p^k*q^{n-k},Pn(k)=Cn(k)∗p  

k

∗q  

n−k

,

then the probability that the ruling party will win 0 rounds is:

P4(0) = q^4 = 0.4^4 = 0.0256;P4(0)=q  

4

=0.4  

4

=0.0256;

the probability that the ruling party will win 1 round is:

P4(1) = 4!/(1!\times3!)\times0.6^1\times0.4^3 = 0.1536;P4(1)=4!/(1!×3!)×0.6  

1

×0.4  

3

=0.1536;

the probability that the ruling party will win 2 rounds is:

P4(2) = 4!/(2!\times2!)\times0.6^2\times0.4^2 = 0.3456;P4(2)=4!/(2!×2!)×0.6  

2

×0.4  

2

=0.3456;

the probability that the ruling party will win 3 rounds is:

P4(3) = 4!/(3!\times1!)\times0.6^3\times0.4^1 = 0.3456;P4(3)=4!/(3!×1!)×0.6  

3

×0.4  

1

=0.3456;

the probability that the ruling party will win all 4 rounds is:

P4(4) = 0.6^4 = 0.1296.P4(4)=0.6  

4

=0.1296.

b.The probability that the ruling party will win at least 1 round is:

P(1, 2, 3, 4) = 1 - q^4 = 1- 0.0256 = 0.9744.P(1,2,3,4)=1−q  

4

=1−0.0256=0.9744.