SOLUTION: there are 3 boxes containing respectively 1 white, 2 red, 3 black balls; 2 white, 3 red, 1 black ball; 3 white, 1 red and 2 black balls. a box is chosen at random and from it two b

Algebra ->  Probability-and-statistics -> SOLUTION: there are 3 boxes containing respectively 1 white, 2 red, 3 black balls; 2 white, 3 red, 1 black ball; 3 white, 1 red and 2 black balls. a box is chosen at random and from it two b      Log On


   



Question 585040: there are 3 boxes containing respectively 1 white, 2 red, 3 black balls; 2 white, 3 red, 1 black ball; 3 white, 1 red and 2 black balls. a box is chosen at random and from it two balls are drawn at random. the two balls are 1 red and 1 white. what is the probability that they come from (1) first box (11) second box (111) third box?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
you can draw a table as shown below:
                W      R     B
box number 1    1      2     3
box number 2    2      3     1
box number 3    3      1     2

you take a box at random and draw 2 balls.
the 2 balls are 1 red and 1 white.

the probability that the balls came from box number 1 is:
1/6 * 2/5 = 2/30 * 2 possible ways this can happen = 4/30 = 2/15
the probability that the balls came from box number 2 is:
2/6 * 3/5 = 6/30 * 2 possible ways this can happen = 12/30 = 6/15
the probability that the balls came from box number 3 is:
3/6 * 1/5 = 3/30 * 2 possible ways this can happen = 6/30 = 3/15

you can also look at this as the possible ways you can get each type ball out of the total possible ways you can get 2 balls out of 6.
this would use the combination formula of nCx where n is the total number of selections and x is the number you want to draw from.

using this type of analysis, you get:
the probability that the balls came from box number 1 is:
(1C1 * 2C1) / 6C2) = (1 * 2) / (15) = 2/15
the probability that the balls came from box number 2 is:
(2C1 * 3C1) / (6C2) = (2 * 3) / (15) = 6/15
the probability that the balls came from box number 3 is:
(3C1 * 1C1) / (6C2) = (3 * 1) / (15) = 3/15





Answer by ikleyn(52899) About Me  (Show Source):
You can put this solution on YOUR website!
.
There are 3 boxes containing respectively 1 white, 2 red, 3 black balls; 
                                          2 white, 3 red, 1 black ball; 
                                          3 white, 1 red and 2 black balls. 
a box is chosen at random and from it two balls are drawn at random. the two balls are 1 red and 1 white. 
what is the probability that they come from (1) first box (11) second box (111) third box?
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        I think that the meaning of this problem is different from that as interpreted by tutor @Theo.
        I will solve it in my interpretation.


If you randomly select box #1, then the probability to get 1 red and 1 white ball is

    2%2A%281%2F6%29%2A%281%2F5%29 = 4%2F30 = 2%2F15.


If you randomly select box #2, then the probability to get 1 red and 1 white ball is

    2%2A%282%2F6%29%2A%283%2F5%29 = 12%2F30 = 6%2F15.


If you randomly select box #3, then the probability to get 1 red and 1 white ball is

    2%2A%283%2F6%29%2A%281%2F5%29 = 6%2F30 = 3%2F15.



The probability to get 1 red and 1 white ball as the outcome from this experiment is

    %281%2F3%29%2A%282%2F15%29 + %281%2F3%29%2A%286%2F15%29 + %281%2F3%29%2A%283%2F15%29 = %282%2B6%2B3%29%2F%283%2A15%29 = 11%2F45.



Now we apply the formula for conditional probability

    P(from box 1) = %281%2F3%29 . ( %28%282%2F15%29%29%2F%28%2811%2F45%29%29 ) = %281%2F3%29 . %282%2A3%29%2F11 = 2%2F11.


    P(from box 2) = %281%2F3%29 . ( %28%286%2F15%29%29%2F%28%2811%2F45%29%29 ) = %281%2F3%29 . %286%2A3%29%2F11 = 6%2F11.


    P(from box 3) = %281%2F3%29 . ( %28%283%2F15%29%29%2F%28%2811%2F45%29%29 ) = %281%2F3%29 . %283%2A3%29%2F11 = 3%2F11.


ANSWER.  P(from box 1) = 2%2F11.  P(from box 2) = 6%2F11.  P(from box 3) = 3%2F11.