Answer:
(a) [tex]Probability = 0.7599[/tex]
(b) [tex]Probability = 0.2646[/tex]
Explanation:
Represent losing with L and winning with W.
So:
[tex]L = 0.7[/tex] --- Given
[tex]n = 4[/tex]
Probability of winning would be:
[tex]W = 1 - L[/tex]
[tex]W = 1 - 0.7[/tex]
[tex]W = 0.3[/tex]
The question illustrates binomial probability and will be solved using the following binomial expansion;
[tex](L + W)^4 = L^4 + 4L^3W + 6L^2W^2 + 4LW^3 + W^4[/tex]
So:
Solving (a): Winning at least 1
We look at the above and we list out the terms where the powers of W is at least 1; i.e., 1,2,3 and 4
So, we have:
[tex]Probability = 4L^3W + 6L^2W^2 + 4LW^3 + W^4[/tex]
Substitute value for W and L
[tex]Probability = 4 * 0.7^3*0.3 + 6*0.7^2*0.3^2 + 4*0.7*0.3^3 + 0.3^4[/tex]
[tex]Probability = 0.7599[/tex]
Hence, the probability of her winning at least one is 0.7599
Solving (a): Wining exactly 2
We look at the above and we list out the terms where the powers of W is exactly 2
So, we have:
[tex]Probability = 6L^2W^2[/tex]
Substitute value for W and L
[tex]Probability = 6*0.7^2*0.3^2[/tex]
[tex]Probability = 0.2646[/tex]
Hence, the probability of her winning exactly two is 0.2646
