SOLUTION: 51. Gendar in Sequences of babies. Assuming boy and girl are equally likely, find the probability that would take.
A. at least 3 births to obtain 2 girls.
B. at least 4 births to
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Question 206965This question is from textbook mathematical ideas
: 51. Gendar in Sequences of babies. Assuming boy and girl are equally likely, find the probability that would take.
A. at least 3 births to obtain 2 girls.
B. at least 4 births to obtain 2 girls.
C. at least 5 births to obtain 2 girls.
52.Drawing Cards. Cards are drawn, without replacement, from an ordinary 52-card deck.
A. How many must be drawn before the probability of obtaining at least one face card is greater than 1/2?
B. How many must be drawn before the probability of obtaining at least one king is greater than 1/2?
This question is from textbook mathematical ideas
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Gender in Sequences of babies. Assuming boy and girl are equally likely, find the probability that would take.
A. at least 3 births to obtain 2 girls.
3C2(1/2)^2(1/2)^3 = 3/32
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B. at least 4 births to obtain 2 girls.
4C2(1/2)^2(1/2)^2 = 6/64 = 3/32
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C. at least 5 births to obtain 2 girls.
5C2(1/2)^2(1/2)^3 = 10/128
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52.Drawing Cards. Cards are drawn, without replacement, from an ordinary 52-card deck.
A. How many must be drawn before the probability of obtaining at least one face card is greater than 1/2?
Without replacement the probability is never greater than 1/2.
Prob on 1st draw is 12/52 = 0.2308
Prob on the 2nd is (12/51)(40/52)= 0.1810
and the Probabilities get smaller as you go further.
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B. How many must be drawn before the probability of obtaining at least one king is greater than 1/2?
If it's without replacement the answer is the same.
With replacement it looks like this:
52Cn(12/52)^n(40/52)^(n-1)
When n=8 the probability is 0.0587
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Cheers,
Stan H.
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