Question

(1 point) Urn A has 5 white and 17 red balls. Urn B has 9 white and 12 red balls. We flip a fair coin. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Suppose that a white ball is selected. What is the probability that the coin landed heads

Answer 1

Answer:

The probability that the coin landed heads is 65.3%.

Step-by-step explanation:

Given : Urn A has 5 white and 17 red balls. Urn B has 9 white and 12 red balls. We flip a fair coin. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Suppose that a white ball is selected.

To find : What is the probability that the coin landed heads ?

Solution :

Let the event A be the ball taken from Urn A (5 white and 17 red balls)

Let B=A'- the ball taken from urn B(9 white and 12 red balls)

Let W be event that a white ball is selected.

An urn is chosen based on a toss of a fair coin.

P(A) = coin landed on heads = $\frac{1}{2}$

P(B) = coin landed on tails = $P(A')=\frac{1}{2}$

$P(W/A)=\frac{5}{22}$ and $P(W/B)=\frac{3}{7}$

Using Bayes formula,

$P(B/W)=\dfrac{P(W/B)P(B)}{P(W/B)P(B)+P(W/A)P(A)}$

$P(B/W)=\dfrac{\frac{3}{7}\times \frac{1}{2}}{\frac{3}{7}\times \frac{1}{2}+\frac{5}{22}\times \frac{1}{2}}$

$P(B/W)=\dfrac{\frac{3}{14}}{\frac{3}{14}+\frac{5}{44}}$

$P(B/W)=\dfrac{\frac{3}{14}}{\frac{132+70}{14\times 44}}$

$P(B/W)=\dfrac{\frac{3}{14}}{\frac{202}{616}}$

$P(B/W)=\dfrac{3\times 616}{14\times 202}$

$P(B/W)=\dfrac{66}{101}$

$P(B/W)=65.3\%$

Therefore, the probability that the coin landed heads is 65.3%.