Two students, Alice and Bob, forgot to put their names on their exam papers. The professor knows that Alice does well with probability 0.8, and Bob does well with probability 0.4, independently of each other. What is the probability that exactly one of Alice and Bob does well on the exam

Answer:

[tex]Probability = 0.56[/tex]

Step-by-step explanation:

Represent the probability that Alice does well with P(A), the probability that Bob does well with P(B).

So, we have:

[tex]P(A) = 0.8[/tex]

[tex]P(B) = 0.4[/tex]

Required

Determine the probability that only one of them do well

This event is represented as: (A and B') or (B and A')

Where B' means Bob does not perform well and A' means Alice does not perform well.

And:

[tex]P(B') = 1 - P(B)[/tex]

[tex]P(A') = 1 - P(A)[/tex]

So, the probability becomes:

[tex]Probability = P(A\ and\ B')\ or\ P(B\ and\ A')[/tex]

[tex]Probability = P(A) * P(B')\ +\ P(B)*P(A')[/tex]

[tex]Probability = P(A) * (1-P(B)) +\ P(B)*(1-P(A))[/tex]

[tex]Probability = 0.8 * (1-0.4) +\ 0.4*(1-0.8)[/tex]

[tex]Probability = 0.8 * 0.6 +\ 0.4*0.2[/tex]

[tex]Probability = 0.48 + 0.08[/tex]

[tex]Probability = 0.56[/tex]

Hence, the required probability is 0.56

so alot of the text got cut off so im going to re wright the problem

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