# What is the probability that two red balls are chosen. Also,given that two red balls are obtained,find the conditional probability that a 1 or 6 was rolled on the dieBag A contains 2 red balls and...

What is the probability that two red balls are chosen. Also,given that two red balls are obtained,find the conditional probability that a 1 or 6 was rolled on the die

Bag A contains 2 red balls and 3 green balls.Two balls are chosen at random from the bag without replacement.Let X denote the number of red balls chosen.The probability distribution for X. When X=0,1,2 then P(X=x) is 3/10, 6/10, and 1/10.

Bag B contains 4 red balls and 2 green balls.Two balls are chosen at random from bag B.

A standard die with six faces is rolled.If a 1 or 6 is obtained,two balls are chosen from bag A,otherwise two balls are chosen from bag B.

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You need to evaluate the probability that two red balls are chosen such that:

`P(1 or 6) = P(1) + P(6)`

`P(1 or 6) = 1/6 + 1/6`

`P(1 or 6) = 2/6`

The problem provides the information that if x = 2, then `P(X=2) = 1/10` such that:

`P(A nn R R) = 2/6*1/10 = 2/60`

The problem provides the information that the bag B contains 4 red balls and 2 green balls and two balls are chosen at random from bag B.

The probability to randomly select two red balls from bag B is `4/6*3/5 = 12/30` .

The probability to randomly select one red ball and one green ball from bag B is `4/6*2/5 = 8/30` .

`P(B nn R R) = 4/6*3/5 = 12/30`

Since the probability 1 or 6 comes from bag A, then you need to evaluate `P(A) = (2/60)/(P(X=0)) =gt P(A) = (2/60)/(3/10) = 2/18 = 1/9.`

**Hence, evaluating the probability under given conditions yields `P(A) = 1/9.` **