Question

Alison has all her money invested in two mutual funds, a and
b. she knows that there is a 40% chance that fund a will rise in price, and a 60% chance that fund b will rise in price given that fund a rises in price. there is also a 20% chance that fund b will rise in price. what is the probability that at least one of the funds will rise in price?

Answer 1

The probability that at least one of the funds will rise will be given by:
P(a or b)=P(a)+P(b)-P(a and b)
Probability that a occurs will be:
P(a)=0.4
Probability that b occurs will be:
 P(b)=0.6
Probability that both events occur was:
P(a and b)=0.2
therefore:
Probability that at least one occurs will be:
P(a or b)=0.4+0.6-0.2
=1-0.2
=0.8
thus
P(a or b)=0.8

Answer 2

Answer: The probability  that at least one of the funds will rise in price is 0.36.

Let P(A) be the probability that fund A will rise.

Let P(B) be the probability that fund B will rise.

We have

P(A) = 0.40

P(B) =0.20

P(B given A) = 0.60

The formula for conditional probability is given as :

$\textbf{P(B given A) = \frac{P(A and B))}{P(A)}}$

From the equation above we can derive P(A and B) as

$\textbf{P(B given A)* P(A) = P(A and B)}$

Substituting the values we get,

$\textbf{0.60* 0.4 = P(A and B)}$

P(A and B) = 0.24.

The P(A or B) will give us the probability that at least one of the funds increase in price.

Since the P(A) and P(B) are not mutually exclusive events, we use the following formula to find the P(A or B)

$\textbf{P(A or B)= P(A) + P(B) - P(A\cap B)}$

where

$P(A\cap B)$ is P(A and B)

Substituting the values we get,

$\textbf{P(A or B)= 0.40 + 0.20 - 0.24 = 0.36}$