Question 1119914: Please help me solve this in simple way. Thank You.
Approximately 14% of people are left-handed. If two people are selected at random, what is the probability of the following events?
A. P(Both are right-handed) =
B. P(Both are left-handed) =
C. P(One is right-handed and the other is left-handed) =
D. P(At least one is right-handed) =
Answer by ikleyn(52794)   (Show Source):
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Please help me solve this in simple way. Thank You.
Approximately 14% of people are left-handed. If two people are selected at random, what is the probability of the following events?
A. P(Both are right-handed) = P(R,R) = P(R)*P(R) = (1-0.14)*(1-0.14) = 0.86*0.86 = 0.7396.
B. P(Both are left-handed) = P(L,L) = P(L)*P(L) = 0.14*0.14 = 0.0196.
C. P(One is right-handed and the other is left-handed) = P(R,L) = P(R)*P(L) = 0.86*0.14 = 0.1204.
D. P(At least one is right-handed) = Complement to bProbability = 1 - P(L)*P(L) = 1 - 0.0196 = 0.9804.
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