Events A1, A2 and A3 form a partiton of the sample space S with probabilities P(A1)=0.1, P(A2)=0.1, P(A3)=0.8.

If E is an event in S with P(E|A1)=0.3, P(E|A2)=0.3, P(E|A3)=0.6, compute
P(E)=
P(A1|E)=
P(A2|E)=
P(A3|E)=

To compute the probability of event E, we can use the law of total probability. According to this law, we can calculate P(E) by summing the probabilities of E occurring given each individual event in the partition, weighted by their respective probabilities.

So, we have:

P(E) = P(E|A1) * P(A1) + P(E|A2) * P(A2) + P(E|A3) * P(A3)

Substituting the given values:

P(E) = 0.3 * 0.1 + 0.3 * 0.1 + 0.6 * 0.8
= 0.03 + 0.03 + 0.48
= 0.54

Thus, the probability of event E occurring is 0.54.

To compute P(A1|E), we can use Bayes' theorem, which states that:

P(A|B) = (P(B|A) * P(A)) / P(B)

Applying this to our problem and substituting the given values:

P(A1|E) = (P(E|A1) * P(A1)) / P(E)

P(A1|E) = (0.3 * 0.1) / 0.54
= 0.03 / 0.54
= 0.0556 (approximately)

Therefore, the probability of event A1 occurring given that event E has occurred is approximately 0.0556.

Similarly, we can calculate P(A2|E) and P(A3|E) using the same formula and substituting the given values:

P(A2|E) = (P(E|A2) * P(A2)) / P(E)
= (0.3 * 0.1) / 0.54
= 0.03 / 0.54
= 0.0556 (approximately)

P(A3|E) = (P(E|A3) * P(A3)) / P(E)
= (0.6 * 0.8) / 0.54
= 0.48 / 0.54
= 0.8889 (approximately)

Therefore, the probabilities of event A2 occurring given that event E has occurred and event A3 occurring given that event E has occurred are approximately 0.0556 and 0.8889, respectively.