Probability help

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OliverCurtis

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Aug 11, 2020
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Hi Posted the whole question but only need help with part B)
A bag A contains 9 black balls, 6 white balls and 3 red balls. A bag B contains 6 black balls, 2 white balls and 4 green balls. Ali takes out 1 ball from each bag randomly. When Ali takes out 1 ball from one bag, he will put it into the other bag and then takes out one ball from that bag. Find the probability that
(a) the ball is black from bag A, followed by white from bag B,
(b) both the balls are white in colour,
(c) the ball is black or white from bag B, followed by red from bag A,
(d) both the balls are of different colours,
(e) both the balls are not black or white in colours.
The solution they offer is:
(b) Probability of white ball from bag A, followed by white ball from bag B=
\displaystyle=\frac{1}{2}\times\frac{6}{18}\times\frac{3}{13}=\frac{1}{26}

Probability of white from B, followed by white from A=
\displaystyle=\frac{1}{2}\times\frac{2}{12}\times\frac{7}{19}=\frac{7}{228}

Total prob=
\displaystyle\frac{205}{2964}


The 6/18 is obviously the p(white from A) and the 3/13 is p(white from B after A) but what is the 1/2. Theyve done the same for the solution for the other one as well.
Any help would be appreciated

Thanks

Oliver
 
Only half the time does Ali start with bag A.

If he starts with bag B then the chance of picking 1 white from B and putting it into A and getting a white from A is 7/114. Do you see that?
If he starts with bag A then the chance of picking 1 white from A and putting it into B and getting a white from B is 1/13. Do you see that?

Since each bag is equally likely to be chosen first (we must assume this otherwise the problem can not be done) we have the answer being .5(7/114) + .5(1/13).
 
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