Probability basics

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Just a quick Q on the last question (c). Based on research of the average lifetime a couple assumes a probability of $0.75$ that the husband will still be alive in $20$ years while the wife has a chance of $0.8$. How likely are the following events?

(a) Both are still alive in $20$ years.

$$P(A∩B)= 0.75 \cdot 0.8$$

(b) None of them is still alive in $20$ years.

$$P(A∩B)= 0.25 \cdot 0.2$$

(c) At least one of them is still alive in $20$ years.

Now for c I used

$$P(A∪B)= 0.75 + 0.8 -(0.75\cdot0.8)$$

but my friend said it's wrong and said I should use $(A∩B)$ but with additions. Is this correct? Was I wrong?

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1
On BEST ANSWER

You're correct and your friend is incorrect.

Alternatively for $(c)$ you could have done

$$P(A\cup B)=(0.75\cdot 0.2)+(0.25\cdot 0.8)+(0.75\cdot 0.8)=0.95$$

which agrees with your solution.

One of three things can happen to satisfy $A\cup B$

  • $A$ happens and $B$ doesn't happen

  • $A$ does not happen and $B$ happens

  • $A$ and $B$ both happen
0
On

For c it's just $1- $ the result in b $$P=1-P(A∩B)= 1-0.2 \times 0.25=0.95$$