Bag A contains 2 red balls and 3 green balls. Two balls are chosen at random from the bag without re

Question

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 following table shows the probability distribution for $X$. $X$ 0 1 2 $P(X=x)$ $\frac3{10}$ $\frac6{10}$ $\frac1{10}$ Calculate $E(X)$, the mean number of red balls chosen. Bag B contains 4 red balls and 2 green balls. Two balls are chosen at random from bag B. Draw a tree diagram to represent the above information, including the probability of each event. Hence find the probability distribution for Y, where Y is the number of red balls chosen. 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. Calculate the probability that two red balls are chosen. Given that two red balls are obtained, find the conditional probability that a 1 or 6 was rolled on the die.

Accepted Answer

$P(\text{two red})=\frac13(\frac1{10})+\frac23(\frac25)=\frac3{10}$. Given two red, $P(1\text{ or }6)=\frac{(1/3)(1/10)}{3/10}=\frac19$.

Use without-replacement distributions, a tree, total probability and Bayes conditional probability.

Use without-replacement distributions, a tree, total probability and Bayes conditional probability.

Solution:

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