GO Classes 2024 | Weekly Quiz 8 | Conditional Probability | Question: 5

Answer 1

Let A be the event that Alice can solve the puzzle. B be the event that Bob can solve the puzzle. And C be the event that Carl can solve the puzzle.

We can imply from the question that $A$, $B$, and $C$ are Independent events. Because each of those events does not affect the probability of happening of other events. Here,  $A^{c}$, $B^{c}$, and $C^{c}$ are also independent events for the same reason.

$p(A^{c}\cap B^{c}\cap C^{c}) = p(A^{c}) * p(B^{c}) * p(C^{c}) = 0.3 * 0.4 * 0.15 = 0.018$

Answer 2

Let  A=Alice can solve the puzzle without making a mistake.     P(A)=0.7

B=Bob can solve the puzzle without making a mistake.              P(B)=0.6

C=Carl can solve the puzzle without making  a mistake.             P(C)=0.85

From the given information we can state that if A happens then there is no effect on probabilities on B and C ,Same thing if B happens and also C happens.Means A,B,C are independent events.Now,

P(A’∩B’∩C’)=P(A’)*P(B’)*P(C’)={1-0.7}*{1-0.6}*{1-0.85}=0.3*0.4*0.15=0.018