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There are 3 red and 7 black balls in first pot. There are 41 red and 59 black balls in second pot. And there are 481 red and 519 black balls in third pot. One of the three pots is chosen according to the random principle and after that one ball is blindly chosen. How big is the probability to get a red ball?

Solution: $\frac{397}{1000}$

How big is the probability of taking a red ball from a fourth pot, when previously the content of the first three pots is placed in empty fourth pot?

Solution: $\frac{35}{74}$

My idea:

I was thinking to use formula: $P(A|B)=\frac{P(B|A)P(A)}{P(B)} $ But I am not sure how to connect this formula with the things that are wanted in this example.

I calculated only: $P(Red)=\frac{3}{10}$ and $P(Black)=\frac{7}{10}$

2 Answers 2

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Why do you want to use conditional probability? Just use simple probability and the rule of probabilistic multiplication for actions done in a sequence.

P(R1)=3/(3+7) P(R2)=41/(41+59) P(R3)=481/(519)

P(choose_the_pot and red ball) 

                           = P(choose_the_pot) * P(red ball in that pot)

                           = 1/3 * P(R1 or R2 or R3)

                           = 1/3 * (P(R1)+P(R2)+P(R3)) = 397/1000

P(R4)= (3+41+481) / (10+100+1000) = 35/74

Simple!

2

Hints:

For part $(a) $, there is no need to use Bayes theorem. The probability can be calculated by adding the individual ones, that is, $$P_1 =\frac {1}{3}\times \frac {3}{10} + \frac {1}{3}\times \frac {41}{100} + \frac {1}{3}\times \frac {481}{1000}$$

For part $(b) $, there are a total of $1110$ balls, with $525$ red ones and the remaining black as all the contents are emptied in a new one.

Hope you can take it from here.