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Homework Help: Conditional probability questions ?

  1. Apr 17, 2006 #1
    Please help me to solve the following questions :
    1) There are three box : box X has 10 bulbs which 4 are defective
    Box Y has 6 bulbs which 1 are defective
    Box Z has 8 bulbs which 3 are defective
    a box is chosen at random and the buld is randomly selected from the choosen box . If the bld is not defective, find the probabily that it come from box Z .(Answer = 75/247)

    2) A pair of fair dice is thrown . If two numbers are different , find the probability that
    a)the sum is 6( answer=2/15)
    b)an ace appears (answer=1/3)
    c)the sum is 4 or less (answer = 2/15)
    d)the sum is even (answer is 2/5)
    the sum exceeds nine (answer is 2/15)

    3)One couple has two children , find the probability both children are boys if
    a)at least one of the children is a boy (answer is 1/3)
    b)the older child is a boy (answer is 1/2)

    I cannot solve most of the questions above because I don't know how to find the P(A n B) since P(A|B)=P(A n B)/P(B) . So, please help me to solve the questions .Thank you .
  2. jcsd
  3. Apr 17, 2006 #2


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    If P(A|B)= P(A n B)/P(B) then P(A n B)= P(A|B)P(B)! In particular, if
    either event A or B must happen, and A and B are mutually exclusive, then
    P(C)= P(C|A)P(A)+ P(C|B)P(B).

    Let N be the event that a non-defective bulb is chosen. Then
    P(N)= P(N|X)P(X)+ P(N|Y)P(Y)+ P(N|Z)P(Z). We can also see that, since P(N|Z)P(Z)= P(Z|N)P(N),
    P(Z|N)= P(N|Z)P(Z)/P(N)= P(N|Z)P(Z)/(P(N|X)P(X)+ P(N|Y)P(Y)+ P(N|Z)P(Z)) where all of the probabilities on the right are "a-priori"- before we select a bulb.
    There are 3 boxs and we choose them at random so a-priori P(X)= P(Y)= P(Z)= 1/3. X has 10 bulbs of which 6 are non-defective so P(N|X)= 6/10= 3/5. Y has 6 bulbs of which 5 are non-defective so P(N|Y)= 5/6. Z has 8 bulbs of which 5 are non-defective so P(N|Z)= 5/8. That is,
    P(N|X)P(X)+ P(N|Y)P(Y)+ P(N|Z)P(Z)= (3/5)(1/3)+ (5/6)(1/3)+ (5/8)(1/3)= 247/360.
    Again, P(N|Z)= 5/8 and P(Z)= 1/3 so P(Z|N)= P(N|Z)P(Z)/(P(N|X)P(X)+ P(N|Y)P(Y)+ P(N|Z)P(Z)) becomes P(Z|N)= (5/8)(1/3)/(247/360)=
    (5/24)(360/247)= (5)(15)/247= 75/247.

    You know that there are 6*6= 36 "equally likely combinations, 6 of which have both numbers the same so 30 have both numbers different. Now,
    (a) how many of those add to 6? ((1,5),(2, 4), (4,2), (1,5)).
    (b) how many of those have an ace?
    (c) how many of those sum to 4 or less?
    (d) how many of those have even sum?
    (e?) how may of those have sum greater than 9?

    Surprising that those are different answers, isn't it?
    Assuming birth of boy or girl is equally likely, there are four equally likely ways to have two children. (Boy, Boy), (Boy, Girl), (Girl, Boy), (Girl, Girl) where they are listed in order of birth.
    If "at least one of the children is a boy", the we can ignore (Girl, Girl) and have only (Boy, Boy), (Boy, Girl), (Girl, Boy). From that information, what is the probability of (Boy, Boy)?
    If "the older child is a boy, then we can ignore both (Girl, Girl) and (Girl, Boy). Now what is the probability of (Boy, Boy)?
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