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# Probability

last edited by 16 years, 4 months ago

# Problem 1 of 3

It is known that 53% of graduating students are boys. Three grads are chosen at random. Given that at least two of the three grads are boys, determine the probability that all three of the grads are boys. (Answer accurate to at least 4 decimal places.)

Solution

P(B)=0.53

P(6)=0.47

P(3B/at least 2B)=(3B)/((P(3B))+(P(2B)))

P(3B)=(0.53)^3=0.148877

P(2B)=3!/2!(o.53)^2(0.47)=0.396069

=.148877/(0.148877+0.396069)

=0.148877/0.544964

=0.693823

# Problem 2 of 3

A biased (weighted) coin is designed so that the probability of a head on each flip is (3/5).

(a) If this biased coin is flipped 3 times, what is the probability that the first 2 flips are tails and the third flip is a head?

(b) If this biased coin is flipped until exactly 2 heads appear, what is the probability that it takes exactly 3 flips until the second head appears?

(c) If this biased coin is flipped 7 times, what is the probability that exactly 3 or 4 heads appear?

Solution

(a)

P(tails)= (2/5)

(b)

P(h,t,h) or P(t,h,h)

P(hth) = P(3/5)(2/5)(3/5) = 0.144(14.4%)

or

P(thh) = (2/5)(3/5)(3/5) = 0.144(14.4%)

(c)

P(4h,3t) = (7!)/(4!)(3!) * (3/5)^4 (2/5)^3 = 0.290304(29%)

P(3h,4t) = (7!)/(3!)(4!) * (/5)^3 (2/5)^4 = 0.193536(19.353%)

therefore the probability that 3 or 4 heads will appear of flipped 7 times is

(0.29) + (0.193536) = .483536(48%)

# Problem 3 of 3

A new lie detector test has been devised and must be tested before it is put into use. One hundred people are selected at random and each person draws a card from a box of 100. Half the cards instruct the person to lie and the other half instruct the person to tell the truth. The new lie detector test indicates lying in 80% of those who did lie and 5% of those who told the truth.

(a) What is the probability that a randomly chosen person has really lied, given that the test indicates lying?

(b) What is the probability that a person actually told the truth if the test indicates lying?

Solution