Question

g An urn contains 10 balls, of which 3 are red, 2 are yellow, and 5 are blue. Five balls are randomly selected one at a time, replacing each ball after selecting it. What is the probability that fewer than 2 of the selected balls are red given that exactly 2 of the selected balls are yellow? Give your answer as a decimal rounded to four places (i.e. X.XXXX)

Answer 1

Answer:

The probability that fewer than 2 of the selected balls are red given that exactly 2 of the selected balls are yellow is 0.7840.

Step-by-step explanation:

The balls in the urn are:

R = red balls = 3

Y = yellow balls = 2

B = blue balls = 5

The probability of drawing a red ball is, $p_{R}=\frac{3}{10}$.

The probability of drawing a yellow ball is, $p_{Y}=\frac{2}{10}$.

The probability of drawing a blue ball is, $p_{B}=\frac{5}{10}$.

Five balls are drawn from the urn with replacement.

The conditional event of selecting less than 2 red balls given that exactly 2 yellow balls have already been selected is same as selecting less than 2 red balls in 3 draws.

The event of the number of red balls selected follows a binomial distribution with parameters n = 3 and $p_{R}=\frac{3}{10}$.

The probability mass function of a Binomial distribution is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3....$

Compute the probability of selecting less than 2 red balls in 3 draws as follows:

P (R < 2) = P (R = 0) + P (R = 1)

              $={3\choose 0}(\frac{3}{10})^{0}(1-\frac{3}{10})^{3-0}+{3\choose 1}(\frac{3}{10})^{1}(1-\frac{3}{10})^{3-1}\\=(1\times1\times0.343)+(3\times0.30\times0.49)\\=0.343+0.441\\=0.784$

Thus, the probability that fewer than 2 of the selected balls are red given that exactly 2 of the selected balls are yellow is 0.7840.