Consider a family with 4 children. Assume the probability that one child is a bo

Question

Consider a family with 4 children. Assume the probability that one child is a boy is 0.5 and the probability that one child is a girl is also 0.5, and that the events "boy" and "girl" are independent.

(a) List the equally likely events for the gender of the 4 children, from oldest to youngest. (Let M represent a boy (male) and F represent a girl (female). Select all that apply.)

MFFM FFMM three M's, one F FFFM MMFF MMMF FMMF MMFM FMFF MFMM two M's, two F's FFFF MFMF FMFM FMMM MMMM FFMF MFFF one M, three F's

Notice that the complement of the event "all four children are male" is "at least one of the children is female." Use this information to compute the probability that at least one child is female. (Enter your answer as a fraction.)

Explanation / Answer

Since the event that a child is Girl or Boy is independent, each of the children can be either a girl or a boy ( i.e. 2 possibilities)

Thus, for 4 childrens total number of possible cases can be 2^4 = 2×2×2×2 = 16 ( Beacuse each of the children from oldest to youngest can be boy or girl)

Thus, the equally likely events are:

MMMM, MMMF, MMFM, MFMM, FMMM, MMFF, MFMF, MFFM, FMFM, FFMM, FMMF, MFFF, FMFF, FFMF, FFFM, FFFF

( a total of 16)

The probability of occorunce of all of these events = 1/16

Now, probability that at least one child is female

= 1 - Probability that no child is female

= 1 - 1/16 = 15/16

(Since only the event MMMM out of these 16 events does not have a female child)

Thus, the required probability is =( 15/16) or 0.9375

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