Maths....Question......急

2007-12-02 11:20 pm

1.Mrs.Ho.has.3.sons.and.3.daughters.while.Mrs.Chan.has.2.sons.and..daughters.If.a.child.is.chosen.randomly.from.each.family,find.the.probability.that.both.children.are.of.the.same.sex.

2.Stephen.answers.four.multiple.four.multiple.choice.questions.by.wild.guessing.If.each.question.contains.four.choices.in.which.only.one.is.correct,find.the.probability.that.Stephen.gets
[a].all.answers.correct
[b].only.three.answers.correct

更新1:

1.Mrs.Ho.has.3.sons.and.3.daughters.while.Mrs.Chan.has.2.sons.and.4.daughters. 2.Stephen.answers.four.multiple.choice.questions.by.wild.guessing.

更新2:

但係2b個答案係3/64

回答 (2)

用戶46949

2007-12-03 5:14 am

✔ 最佳答案

1. p(both are daughters) + p(both are sons)
=3/6 x 4/6 + 3/6 x 2/6
=1/3 + 1/6
=2/6+1/6
=1/2 //
=0.5//

2a p(all are correct)
=(1/4)^4
=1/256 //

2b p(3 answers correct)
=(1/4)^3 (3/4) x 4
=3/64 //

p(TTTF) +P(TTFT) + P(TFTT) +P(FTTT)
// there are four possibility satisfied the requirement)
= (1/4)(1/4)(1/4)(3/4) x 4
=3/64//

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Cesc

2007-12-02 11:26 pm

1. P= 3/6 x 2/4 + 3/6x2/4= 1/2

2a P= (1/4)^4 = 1/256

b. P(1/4)^3x(1/4)x4=1/64

參考: me

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