Question

three turns of the same appearance are given as follows urn A contains 5 red and 6 white balls ,urn B contains 6 red and 4 white balls,urn C contains 3 red and 5 white balls A urn is selected at random and ball is drawn from the urn (A) what is probability that ball drawn is (B)what is probability that ball is from urn A given that ball is red

Answer 1

Answer on Question #81713 – Math – Statistics and Probability

Three turns of the same appearance are given as follows: urn A contains 5 red and 6 white balls, urn B contains 6 red and 4 white balls, urn C contains 3 red and 5 white balls. An urn is selected at random and ball is drawn from the urn.

Question

(A) What is probability that ball drawn is red, white?

Solution

Let $E_A, E_B, E_C$ and $R$ be the events defined as follows:

$E_A = \text{urn A is chosen}, E_B = \text{urn B is chosen}, E_C = \text{urn C is chosen},$ and $R = \text{ball drawn is red}.$

Since there are three urns of the same appearance and one of three urns is chosen at random, therefore

$$P(E_A) = P(E_B) = P(E_C) = \frac{1}{3}$$

The probability of drawing a red ball from urn A is

$$P(R|E_A) = \frac{5}{5 + 6} = \frac{5}{11}$$

The probability of drawing a red ball from urn B is

$$P(R|E_B) = \frac{6}{6 + 4} = \frac{3}{5}$$

The probability of drawing a red ball from urn C is

$$P(R|E_C) = \frac{3}{3 + 5} = \frac{3}{8}$$

We are required to find $P(R)$.

By Law of total probability we have

$$P(R) = P(E_A)P(R|E_A) + P(E_B)P(R|E_B) + P(E_C)P(R|E_C)$$ $$P(R) = \frac{1}{3}\left(\frac{5}{11}\right) + \frac{1}{3}\left(\frac{3}{5}\right) + \frac{1}{3}\left(\frac{3}{8}\right) = \frac{200 + 264 + 165}{1320} = \frac{629}{1320} \approx 0.4765$$

Let $W$ be the event that ball drawn is white.

If we are required to find $P(W)$, then

$$P(W|E_A) = \frac{6}{5 + 6} = \frac{6}{11}$$ $$P(W|E_B) = \frac{4}{6 + 4} = \frac{2}{5}$$ $$P(W|E_C) = \frac{5}{3 + 5} = \frac{5}{8}$$

By Law of total probability we have

$$P(W) = P(E_A)P(W|E_A) + P(E_B)P(W|E_B) + P(E_C)P(W|E_C)$$ $$P(W) = \frac{1}{3} \left(\frac{6}{11}\right) + \frac{1}{3} \left(\frac{2}{5}\right) + \frac{1}{3} \left(\frac{5}{8}\right) = \frac{240 + 176 + 275}{1320} = \frac{691}{1320} \approx 0.5235$$

Check

$$P(R) + P(W) = \frac{629}{1320} + \frac{691}{1320} = 1$$

Question

(B) What is probability that ball is from urn A given that ball is red?

Solution

Let $E_A, E_B, E_C$ and $R$ be the events defined as follows:

$E_A = \text{urn A is chosen}, E_B = \text{urn B is chosen}, E_C = \text{urn C is chosen},$ and

$R = \text{ball drawn is red.}$

Since there are three urns of the same appearance and one of three urns is chosen at random, therefore

$$P(E_A) = P(E_B) = P(E_C) = \frac{1}{3}$$

The probability of drawing a red ball from urn A is

$$P(R|E_A) = \frac{5}{5 + 6} = \frac{5}{11}$$

The probability of drawing a red ball from urn B is

$$P(R|E_B) = \frac{6}{6 + 4} = \frac{3}{5}$$

The probability of drawing a red ball from urn C is

$$P(R|E_C) = \frac{3}{3 + 5} = \frac{3}{8}$$

We are required to find $P(E_A|R)$.

By Bayes' theorem we have

$$P(E_A|R) = \frac{P(E_A)P(R|E_A)}{P(E_A)P(R|E_A) + P(E_B)P(R|E_B) + P(E_C)P(R|E_C)}$$ $$P(E_A|R) = \frac{\frac{1}{3} \left(\frac{5}{11}\right)}{\frac{1}{3} \left(\frac{5}{11}\right) + \frac{1}{3} \left(\frac{3}{5}\right) + \frac{1}{3} \left(\frac{3}{8}\right)} = \frac{200}{200 + 264 + 165} = \frac{200}{629} \approx 0.3180$$

The probability that ball is from urn A given that ball is red

$$P(E_A|R) = \frac{200}{629} \approx 0.3180$$

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