Question

There are three persons A, B and C having different ages. The probability that A survives another 5 years is 0.80, B survives another 5 years is 0.60 and C survives another 5 years is 0.50. The probabilities that A and B survive another 5 years is 0.46, B and C survive another 5 years is 0.32 and A and C survive another 5 years 0.48. The probability that all these three persons survive another 5 years is 0.26. Find the probability that at least one of them survives another 5 years.

• A. 0.8
• B. 0.9
• C. 0.5
• D. 0.25

Answer 1

The correct option is B 0.9

$\text{Given:}P\left(A\right)=0.80P\left(B\right)=0.60P\left(C\right)=0.50P(A\cap B)=0.46P(B\cap C)=0.32P(A\cap C)=0.48P(A\cap B\cap C)=0.26$

The probability that at least one of them survives another 5 years is,
$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)–P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)=0.80+0.60+0.50-0.46-0.32-0.48+0.26=0.90$