Probabilities

1. A family has three children, whatis the probability that there are at least two boys,
given that the first child is a boy?
Sol
A：第1個小孩為boy的事件
B：剩下2個小孩為2 boys的事件
C：剩下2個小孩為1 boy，1 girl的事件
P=P(at least two boys)
=P(three boys)+P(two boys)
=P(AB)+P(AC)
=P(B｜A)*P(A)+P(C｜A)*P(A)
=[(1/2)^2]*1+(1/2)*1
=3/4

2. Bag A contains 7 red balls, and 5 blue balls. Bag B contains 5 redballs and 4
green balls. One ball is randomly drawn from A and put into bag B. Then oneball
is drawn from bag B randomly and put into A.Final, a ball is drawn from A
at random. Find the probability that the final ball drawn from bag A is
(i) blue
(ii) red
Sol
A：第1次從A拿出紅球的事件
B：第1次從A拿出藍球的事件
C：第2次從B拿出紅球的事件
D：第2次從B拿出藍球的事件E：第3次從A拿出紅球的事件
(i)
P=P(E)=P(CE)+P(DE)=P(ACE)+P(BCE)+P(ADE)+P(BDE)
=P(E｜AC)*P(AC)+P(BCE)+P(ADE)+P(BDE)
=P(E｜AC)*P(C｜A)*P(A)+P(BCE)+P(ADE)+P(BDE)
=P(E｜AC)*P(C｜A)*P(A)+ P(E｜BC)*P(C｜B)*P(B)
+P(E｜AD)*P(D｜A)*P(A)+ P(E｜BD)*P(D｜B)*P(B)
=(7/12)*(6/10)*(7/12)+(8/12)*(5/10)*(5/12)
+(6/12)*(4/10)*(7/12)+(7/12)*(5/10)*(5/12)
=(7*6*7+8*5*5+6*4*7+7*5*5)/(12*10*12)
=837/1440
=93/160
(ii) 1－93/160=67/160