SOLUTION: suppose that, of couples having children, in 2 percent both father and mother are left-handed, in 20 percent one is left handed, and in the rest neither is left handed. The chances

Question 763014: suppose that, of couples having children, in 2 percent both father and mother are left-handed, in 20 percent one is left handed, and in the rest neither is left handed. The chances of a child being left-handed are 1 in 2 if both parents are left handed, 1 in 6 if one parent is left handed, and 1 in 16 if neither parents is left handed. What is the probability that neither parent of a left handed child is left handed?
1. What is a probability of a randomly chosen child being left handed? (Hint: the different events are mutually exclusive and we can apply the addition rule for mutually exclusive events)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
suppose that, of couples having children,
in 2 percent both father and mother are left-handed,
in 20 percent one is left handed,
in 78 percent neither is left handed.
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The chances of a child being left-handed are 1 in 2
P(left) = 1/3 if both parents are left handed,
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1 in 6 if one parent is left handed,
P(left) = 1/7
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and 1 in 16 if neither parents is left handed
P(left) = 1/17
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What is the probability that neither parent of a left handed child is left handed?
1. What is a probability of a randomly chosen child being left handed? (Hint: the different events are mutually exclusive and we can apply the addition rule for mutually exclusive events)
Ans: (1/3) + (1/7) + (1/17) = [(7*17) + (3*17) + (3*7)]/[3*7*17] = 89/357
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Cheers,
Stan H.
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