document.write( "Question 1132436: Here’s a question we were unsure of. \r
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document.write( "According to a study of the numbers of male and female children in families in a certain population, a model has been determined that for any subsequent children, the probability that they will be of the same gender as the previous child is 3/5. Using this model, calculate the probability that, in a randomly chosen family of 3 children, there will be;
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document.write( "a) 2 males and 1 female
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document.write( "b) at least one female\r
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document.write( "Hence, construct the probability distribution table of X if X is the number of female selected among the 3 children. \n" );
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Algebra.Com's Answer #749481 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "We need to assume that there is equal probability that the first child is male or female. After that, the probability is 3/5 that the next child will be the same gender and 2/5 that it will be different. Then... \n" ); document.write( "(1) all male \n" ); document.write( "P(MMM) = (1/2)(3/5)(3/5) = 9/50 \n" ); document.write( "(2) 2 male, 1 female \n" ); document.write( "P(MMF) = (1/2)(3/5)(2/5) = 6/50 \n" ); document.write( "P(MFM) = (1/2)(2/5)(2/5) = 4/50 \n" ); document.write( "P(FMM) = (1/2)(2/5)(3/5) = 6/50 \n" ); document.write( "total 16/50 \n" ); document.write( "(3) 1 male, 2 female \n" ); document.write( "P(MFF) = (1/2)(2/5)(3/5) = 6/50 \n" ); document.write( "P(FMF) = (1/2)(2/5)(2/5) = 4/50 \n" ); document.write( "P(FFM) = (1/2)(3/5)(2/5) = 6/50 \n" ); document.write( "total 16/50 \n" ); document.write( "(4) all female \n" ); document.write( "P(FFF) = (1/2)(3/5)(3/5) = 9/50 \n" ); document.write( "Note that the sum of all the probabilities is 1, giving confidence that the calculations were correct. \n" ); document.write( "You can use the probabilities shown to answer the questions that were asked. \n" ); document.write( " |