The probability that a freshman takes English or Math is 0.8. The probability of
ID: 3130320 • Letter: T
Question
The probability that a freshman takes English or Math is 0.8. The probability of taking English is 0.5 and the probability of taking both English and Math is 0.4. What is the probability that a freshman will take Math? A survey shows that for all applicants to a certain college, the probability to be accepted is 0.6. For those accepted, the probability that they will attend the college is 0.8. What is the probability that a randomly selected applicant ends up attending the college? Draw five cards from 52. Find the probabilities that All 5 cards are spades. There are 3 spades and 2 diamonds. All five cards are of the same suit. They contain all 4 aces. One card is drawn from 52. Find the probability that it is an ace given that it is not a king. A box contains 10 light bulbs, 2 of which are defective. Suppose that 3 bulbs are selected in a row, without replacement. Find the probability that All 3 bulbs are good.Explanation / Answer
1.)
P(English U Math) = 0.8
P(English) =0.5
P(English AND Math) = 0.4
P(English U Math) = P(English) + P(Math) - P(English AND Math)
=> 0.8 = 0.5 + P(Math) - 0.4
=> P(Math) = 0.7 Answer
2)
P(Accepted) =0.6
P(Attend College / Accepted) =0.8
P(Attend College / Accepted) = P( Attend College AND Accepted) / P(Accepted)
=> 0.8 = P( Attend College AND Accepted) / 0.6
=>P( Attend College AND Accepted) = 0.48
=> P(Attend College) = 0.48 Answer
Note : A student can only attend if he is accepted , that's why P( Attend College AND Accepted) = P(Attend College)
3)
a. P(All 5 spades) = 13C5 / 52C5 = 0.000495 Answer
b. P(3 spades 2 diamonds ) = (13C3 * 13C2) / 52C5 = 0.00858 Answer
c. P (All 5 same suit) = 4 * 0.000495 = 0.00198 Answer
d. P(Contain all 4 aces) = (52-4)/52C5 = 0.0000184 Answer
4.)
P(Ace/Not King) = P(Ace AND Not King) / P(Not King)
= (4/52) / (48/52)
= 1/12
= 0.0833 Answer
5)
a) P(All 3 bulbs are good) = 1/8 * 1/7 * 1/6 = 0.00297 Answer
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