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it-is-assumed-that-when-children-are-born-they-are-equally-likely-to-be-boys-or-girls-what-is-the-probability-that-a-family-of-four-children-constains-a-three-boys-and-girl-b-two-boys-and-two-girl

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Question Number 140344 by mathdave last updated on 06/May/21
it is assumed that when children are born they are equally likely to be boys or girls. what is the probability that a family of four children constains (a)three boys and girl (b)two boys and two girls Mr W pls help out
$${it}\:{is}\:{assumed}\:{that}\:{when}\:{children}\:{are}\:{born}\:{they}\: \\ $$$${are}\:{equally}\:{likely}\:{to}\:{be}\:{boys}\:{or}\:{girls}. \\ $$$${what}\:{is}\:{the}\:{probability}\:{that}\:{a}\:{family} \\ $$$${of}\:{four}\:{children}\:{constains} \\ $$$$\left({a}\right){three}\:{boys}\:{and}\:{girl} \\ $$$$\left({b}\right){two}\:{boys}\:{and}\:{two}\:{girls} \\ $$$$\:{Mr}\:{W}\:{pls}\:{help}\:{out} \\ $$
Answered by mr W last updated on 06/May/21
(a) probability for a child to be girl is (1/2), probability for one from 4 children to be girl is then p=C_1 ^4 ×(1/2^4 )=(1/4) (b) p=C_2 ^4 ×(1/2^4 )=(3/8)
$$\left({a}\right) \\ $$$${probability}\:{for}\:{a}\:{child}\:{to}\:{be}\:{girl}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}, \\ $$$${probability}\:{for}\:{one}\:{from}\:\mathrm{4}\:{children} \\ $$$${to}\:{be}\:{girl}\:{is}\:{then} \\ $$$${p}={C}_{\mathrm{1}} ^{\mathrm{4}} ×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left({b}\right) \\ $$$${p}={C}_{\mathrm{2}} ^{\mathrm{4}} ×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by mathdave last updated on 06/May/21
thanks so much but pls can u tell me hw u got C_1 ^4 and (1/2^4 ),C_2 ^4 and (1/2^4 )
$${thanks}\:{so}\:{much}\:{but}\:{pls}\:{can}\:{u}\:{tell}\:{me}\:{hw}\:{u}\:{got}\: \\ $$$${C}_{\mathrm{1}} ^{\mathrm{4}} \:{and}\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} },{C}_{\mathrm{2}} ^{\mathrm{4}} \:\:{and}\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} } \\ $$$$ \\ $$
Commented by mr W last updated on 06/May/21
you should read about binomial probability.
$${you}\:{should}\:{read}\:{about}\:{binomial} \\ $$$${probability}. \\ $$
Commented by mr W last updated on 06/May/21
Commented by mr W last updated on 06/May/21
certainly we can also solve in other ways. four different childen, each can be boy or girl, there are totally 2^4 possibilities. (a) one child is girl, there are C_1 ^4 =4 possibilities, i.e. 1 2 3 4 G B B B B G B B B B G B B B B G so the probability is (4/2^4 )=(1/4). (b) two children are boys, there are C_2 ^4 =6 possibilities, i.e. 1 2 3 4 B B G G B G B G B G G B G B B G G B G B G G B B probability is (6/2^4 )=(3/8)
$${certainly}\:{we}\:{can}\:{also}\:{solve}\:{in}\:{other}\: \\ $$$${ways}. \\ $$$$ \\ $$$${four}\:{different}\:{childen},\:{each}\:{can}\:{be} \\ $$$${boy}\:{or}\:{girl},\:{there}\:{are}\:{totally}\:\mathrm{2}^{\mathrm{4}} \: \\ $$$${possibilities}. \\ $$$$\left({a}\right) \\ $$$${one}\:{child}\:{is}\:{girl},\:{there}\:{are}\:{C}_{\mathrm{1}} ^{\mathrm{4}} =\mathrm{4}\: \\ $$$${possibilities},\:{i}.{e}. \\ $$$$\mathrm{1}\:\:\mathrm{2}\:\:\:\mathrm{3}\:\:\:\mathrm{4} \\ $$$${G}\:{B}\:{B}\:{B} \\ $$$${B}\:{G}\:{B}\:{B} \\ $$$${B}\:{B}\:{G}\:{B} \\ $$$${B}\:{B}\:{B}\:{G} \\ $$$${so}\:{the}\:{probability}\:{is}\:\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{4}}. \\ $$$$\left({b}\right) \\ $$$${two}\:{children}\:{are}\:{boys},\:{there}\:{are} \\ $$$${C}_{\mathrm{2}} ^{\mathrm{4}} =\mathrm{6}\:{possibilities},\:{i}.{e}. \\ $$$$\mathrm{1}\:\:\:\mathrm{2}\:\:\mathrm{3}\:\:\mathrm{4} \\ $$$${B}\:{B}\:{G}\:{G} \\ $$$${B}\:{G}\:{B}\:{G} \\ $$$${B}\:{G}\:{G}\:{B} \\ $$$${G}\:{B}\:{B}\:{G} \\ $$$${G}\:{B}\:{G}\:{B} \\ $$$${G}\:{G}\:{B}\:{B} \\ $$$${probability}\:{is}\:\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{4}} }=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by Tawa11 last updated on 06/Sep/21
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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