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If E, F and G are events then P ((E ∪ F)|G) =
- a)P (E|G) + P (G|F) – P ((E ∩ F)|G)
- b)P (E|G) + P (F|G) – P ((E ∩ F)|G)
- c)P (E|G) + P (F|G) – P ((E ∩ F)|F)
- d)P (G|E) + P (F|G) – P ((E ∩ F)|E)
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
If E, F and G are events then P ((E ∪ F)|G) =a)P (E|G) + P (G|F)...
If E, F and G are events then P ((E ∪ F)|G) represents the conditional probability of the given event . therefore P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G) .
Most Upvoted Answer
If E, F and G are events then P ((E ∪ F)|G) =a)P (E|G) + P (G|F)...
To prove that option B is correct, we need to use conditional probability and the properties of set operations.
Let's break down the given expression: P((E ∩ F)|G). This represents the probability of the intersection of events E and F, given that event G has occurred.
Proof:
1. Start with the definition of conditional probability:
P((E ∩ F)|G) = P(E ∩ F ∩ G) / P(G)
2. Expand the intersection using set operations:
P((E ∩ F)|G) = P((E ∩ F) ∩ G) / P(G)
3. Apply the associative property of set intersection:
P((E ∩ F)|G) = P(E ∩ (F ∩ G)) / P(G)
4. Use the conditional probability definition again:
P((E ∩ F)|G) = P(E ∩ (F ∩ G)) / P(G ∩ Ω)
5. Apply the associative property of set intersection again:
P((E ∩ F)|G) = P((E ∩ G) ∩ F) / P(G ∩ Ω)
6. Apply the commutative property of set intersection:
P((E ∩ F)|G) = P((E ∩ G) ∩ F) / P(Ω ∩ G)
7. Use the conditional probability definition one more time:
P((E ∩ F)|G) = P(E ∩ G) ∩ F) / P(G) / P(Ω) / P(G)
8. Simplify the expression:
P((E ∩ F)|G) = P(E|G) * P(F|G) / P(G)
Comparing this result with the options, we can see that option B matches the derived expression: P(E|G) * P(F|G) * P((E ∩ F)|G).
Therefore, the correct answer is option B: P(E|G) * P(F|G) * P((E ∩ F)|G).
Let's break down the given expression: P((E ∩ F)|G). This represents the probability of the intersection of events E and F, given that event G has occurred.
Proof:
1. Start with the definition of conditional probability:
P((E ∩ F)|G) = P(E ∩ F ∩ G) / P(G)
2. Expand the intersection using set operations:
P((E ∩ F)|G) = P((E ∩ F) ∩ G) / P(G)
3. Apply the associative property of set intersection:
P((E ∩ F)|G) = P(E ∩ (F ∩ G)) / P(G)
4. Use the conditional probability definition again:
P((E ∩ F)|G) = P(E ∩ (F ∩ G)) / P(G ∩ Ω)
5. Apply the associative property of set intersection again:
P((E ∩ F)|G) = P((E ∩ G) ∩ F) / P(G ∩ Ω)
6. Apply the commutative property of set intersection:
P((E ∩ F)|G) = P((E ∩ G) ∩ F) / P(Ω ∩ G)
7. Use the conditional probability definition one more time:
P((E ∩ F)|G) = P(E ∩ G) ∩ F) / P(G) / P(Ω) / P(G)
8. Simplify the expression:
P((E ∩ F)|G) = P(E|G) * P(F|G) / P(G)
Comparing this result with the options, we can see that option B matches the derived expression: P(E|G) * P(F|G) * P((E ∩ F)|G).
Therefore, the correct answer is option B: P(E|G) * P(F|G) * P((E ∩ F)|G).
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