A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?
~$3 \over 8$~
~$1 \over 4$~
~$3 \over 16$~
~$1 \over 8$~
~$1 \over 16$~
(2C4)/(2x2x2x2)=6/16=3/8
同上
C4,2/2*2*2*2
C4.2*1/2*1/2*1/2*1/2
生四次,总共有16中可能性。
生两男两女,排列组合;考虑两个女孩或男孩的出生顺序可能性,两个男孩或女孩有一方的顺序决定了,即决定了所有两男两女排列的可能性。
P(2,4)/16, 即3/8.
生4次孩子,每次都有可能生出男或者女,2*2*2*2=16
生出2女2男的可能性:
BBGG
BGBG
BGGB
GGBB
GBGB
GBBG
总6种
所以是6/16=3/8
So here we have four positions and there are two options to fill each position
so total number of cases=2x2x2x2=16
now we need 2boys and 2 girls OR we can say that we simply need 2 boys because if its not a boy it has to be a girl
favourable cases=4C2=6
probability=6/16=3/8
生死次孩子,每次都有可能生出男或者女,2*2*2*2=16
生出2女2男的可能性:
BBGG
BGBG
BGGB
GGBB
GBGB
GBBG
总6种
所以是6/16=3/8
二项式分布C(n,k)*p^k*(1-p)^(n-k)