Find the probability that the new course is introduced.
Question: Three professors A, B and C appear in an interview for the post of Principal. Their chances of getting selected as a principal are 2/9, 4/9, 1/3. The probabilities they introduce new course in the college are 3/10, 1/2, 4/5 respectively. Find the probability that the new course is introduced.
(you are supposed to fill in the blanks for this question.)
Let A, B, C be the events that prof. A, B and C are selected as principal.
Given P(A) = 2/9 , P(B) = 4/9 , P(C) = 1/3 = 3/9
Let N be the event that New Course is introduced
P(N|A) = ___, P(N|B) = ___, P(N|C) = 4/5
N = (A ∩ N) U (B ∩ __) U (__ ∩ N)
P(N) = (A ∩ N) + (B ∩ __) + (__ ∩ N)
= P(A) x P(N|A) + P(__) x ___ + ___ x P(N|C)
= ___ ___ + ___ ___ + ___ ___
= ___ + ___ + ___
= ___
***
Background knowledge/info:
1. Multiplication theorem: The probability of simultaneous occurrence of two events A and B is equal to the product of the probability of the other, given that the first one has occurred.
P(A ∩ B) = P(A) . P(B|A)
P(B|A) = P(A ∩ B) / P(A)
Similarly,
P(A ∩ B) = P(B) . P(A|B)
P(A|B) = P(A ∩ B) / P(B)
2. Independent events: The events that do NOT depend on (or affect the occurrence of) each other. The probability of one event is NOT changed by the realization or outcome of another event.
If A and B are independent events, then
P(A|B) = P(A|B') = P(A)
P(B|A) = P(B|A') = P(B)
P(A ∩ B) = P(A) x P(B)
(you are supposed to fill in the blanks for this question.)
Let A, B, C be the events that prof. A, B and C are selected as principal.
Given P(A) = 2/9 , P(B) = 4/9 , P(C) = 1/3 = 3/9
Let N be the event that New Course is introduced
P(N|A) = ___, P(N|B) = ___, P(N|C) = 4/5
N = (A ∩ N) U (B ∩ __) U (__ ∩ N)
P(N) = (A ∩ N) + (B ∩ __) + (__ ∩ N)
= P(A) x P(N|A) + P(__) x ___ + ___ x P(N|C)
= ___ ___ + ___ ___ + ___ ___
= ___ + ___ + ___
= ___
***
Background knowledge/info:
1. Multiplication theorem: The probability of simultaneous occurrence of two events A and B is equal to the product of the probability of the other, given that the first one has occurred.
P(A ∩ B) = P(A) . P(B|A)
P(B|A) = P(A ∩ B) / P(A)
Similarly,
P(A ∩ B) = P(B) . P(A|B)
P(A|B) = P(A ∩ B) / P(B)
2. Independent events: The events that do NOT depend on (or affect the occurrence of) each other. The probability of one event is NOT changed by the realization or outcome of another event.
If A and B are independent events, then
P(A|B) = P(A|B') = P(A)
P(B|A) = P(B|A') = P(B)
P(A ∩ B) = P(A) x P(B)
1 Answer
Let A, B, C be the events that prof. A, B and C are selected as principal.
Given P(A) = 2/9 , P(B) = 4/9 , P(C) = 1/3 = 3/9
Let N be the event that New Course is introduced
P(N|A) = 3/10, P(N|B) = 1/2, P(N|C) = 4/5
N = (A ∩ N) U (B ∩ N) U (C ∩ N)
P(N) = (A ∩ N) + (B ∩N) + (C ∩ N)
= P(A) x P(N|A) + P(B) x P(N|B) + P(C) x P(N|C)
= 2/9 . 3/10+ 4/9. 1/2 + 3/9.4/5
= 6/90 + 2/9+ 12/45=6/90 + 20/90+ 24/90
= 50/90=5/9.
443
-
Thank you so much, Daniel!!!
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 574 views
- Pro Bono
Related Questions
- What are the odds of at least k same outcomes in n independent trials, each with x equally likely outcomes?
- What is the probability that the last person to board an airplane gets to sit in their assigned seat?
- Probability of more than 1 goal at the end of a match
- Please check if my answers are correct - statistic, probability
- A bag contains 3 red jewels and 7 black jewels. You randomly draw the jewels out one by one without replacement. What is the probability that the last red jewel was the 8th one withdrawn?
- What would be the probability of "breaking the bank" in this 1985 Blackjack game? (Details in body)
- Normal distribution & Probability
- A miner trapped in a mine