Question

if a nd b are two events such ghat p(a) =1/4 p(b)= 1/3 p(aub) =1/2 then p(b/a)​

Answer 1

Answer:

We find that the proportion of spoiled plates is

P(C ∪ G) = P(C) + P(G) − P(C ∩ G)

= 1

10 +

1

10 − 1

50 = 9

50.

The regular inspection passses 9/10 of all plates. Let I be the event of a plate

passing its inspection. The assumption that a plate will certainly pass the

inspection if it is unspoiled is the condition that P(I|Gc ∩Cc) = 1. We are told

that P(I)=9/10 and we also know that P(Cc ∩ Gc)=1 − P(C ∪ G) = 41/50.

From the equation

P(I) = P(I|G ∪ C)P(C ∪ G) + P(I|Gc ∩ Cc

)P(Gc ∩ Cc

),

we find that

P(I|G ∪ C) = P(I) − P(I|Gc ∩ Cc)P(Gc ∩ Cc)

P(G ∪ C)

= 9/10 − 1 × 41/50

9/50 = 4

9

.