Answer:
(a) ![Probability = 0.7599](https://tex.z-dn.net/?f=Probability%20%3D%200.7599)
(b) ![Probability = 0.2646](https://tex.z-dn.net/?f=Probability%20%3D%200.2646)
Explanation:
Represent losing with L and winning with W.
So:
--- Given
![n = 4](https://tex.z-dn.net/?f=n%20%3D%204)
Probability of winning would be:
![W = 1 - L](https://tex.z-dn.net/?f=W%20%3D%201%20-%20L)
![W = 1 - 0.7](https://tex.z-dn.net/?f=W%20%3D%201%20-%200.7)
![W = 0.3](https://tex.z-dn.net/?f=W%20%3D%200.3)
The question illustrates binomial probability and will be solved using the following binomial expansion;
![(L + W)^4 = L^4 + 4L^3W + 6L^2W^2 + 4LW^3 + W^4](https://tex.z-dn.net/?f=%28L%20%2B%20W%29%5E4%20%3D%20L%5E4%20%2B%204L%5E3W%20%2B%206L%5E2W%5E2%20%2B%204LW%5E3%20%2B%20W%5E4)
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:
![Probability = 4L^3W + 6L^2W^2 + 4LW^3 + W^4](https://tex.z-dn.net/?f=Probability%20%3D%204L%5E3W%20%2B%206L%5E2W%5E2%20%2B%204LW%5E3%20%2B%20W%5E4)
Substitute value for W and L
![Probability = 4 * 0.7^3*0.3 + 6*0.7^2*0.3^2 + 4*0.7*0.3^3 + 0.3^4](https://tex.z-dn.net/?f=Probability%20%3D%204%20%2A%200.7%5E3%2A0.3%20%2B%206%2A0.7%5E2%2A0.3%5E2%20%2B%204%2A0.7%2A0.3%5E3%20%2B%200.3%5E4)
![Probability = 0.7599](https://tex.z-dn.net/?f=Probability%20%3D%200.7599)
<em>Hence, the probability of her winning at least one is 0.7599</em>
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:
![Probability = 6L^2W^2](https://tex.z-dn.net/?f=Probability%20%3D%206L%5E2W%5E2)
Substitute value for W and L
![Probability = 6*0.7^2*0.3^2](https://tex.z-dn.net/?f=Probability%20%3D%206%2A0.7%5E2%2A0.3%5E2)
![Probability = 0.2646](https://tex.z-dn.net/?f=Probability%20%3D%200.2646)
<em>Hence, the probability of her winning exactly two is 0.2646</em>