Pool Ball Probability
There lie 4 pool balls on a pool table: two striped and two plain. Two of the pool balls are selected at the same time, at random. Given that one of the selected balls is striped, what's the probability that the other is also striped?
The answer is 0.2.
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2 solutions
We can Solve it Using Baye's Theorem:- P=probability (2 striped)/(probability(2 striped) + probability(1 striped and 1 plain))=(1/6)/[ (1/6) + (4/6)]=1/5 =0.2
Wouldn't the probability be 1/3? This is similar to Bertrand's box paradox in that we are twice as likely to pick the combination with two striped balls, than the combinations with a striped and plain ball, given that we observed a striped ball and that is 100% likely with SS and only 50% likely with SP.
I think that even Bayes' rule shows this: ( S S C o m b i n a t i o n s ) ∗ ( P ( s e e s t r i p e d ∣ S S ) ) + ( S P C o m b . ) ∗ ( P ( s e e s t r i p e d ∣ S P ) ) + ( P P C o m b . ) ∗ ( P ( s e e s t r i p e d ∣ P P ) ) P ( h a v e a n o t h e r s t r i p e d ∣ S S ) = ( 1 ∗ 1 ) + ( 4 ∗ 0 . 5 ) + ( 1 ∗ 0 ) 1 = 3 1
awesome and brilliant!!!!!!!!!!!!!!!!!!!!!!!
Why s1& p2 the order should be important as you said first ball is striped.....so concider the order and your answer will be change
There are four different ways that we can select two pool balls:
Both striped: (1/2)(1/3)=1/6
First striped, second solid: (1/2)(2/3)=2/6
First solid, second striped: (1/2)(2/3)=2/6
Both solid: (1/2)(1/3)=1/6
Since we are given that one of the balls was striped, we ignore scenario 4. The total probability is then the probability of the scenario we want (scenario 1) divided by the sum of the probabilities of all possible scenarios (1, 2, and 3):
(1/6)/((1/6)+(2/6)+(2/6))=1/5
Let the striped balls be denoted by: S 1 and S 2 Let the plain balls be denoted by: P 1 and P 2
There are 6 possible combinations of these 4 balls:
S 1 & S 2
S 1 & P 1
S 1 & P 2
S 2 & P 1
S 2 & P 2
P 1 & P 2
It is given that one of the selected balls is striped, leaving 5 possibilities (highlighted in bold). Out of these 5, only one contains another striped ball, hence the answer is: 1/5 (= 0.2)