Answer :
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, [tex]p_{R}=\frac{3}{10}[/tex].
The probability of drawing a yellow ball is, [tex]p_{Y}=\frac{2}{10}[/tex].
The probability of drawing a blue ball is, [tex]p_{B}=\frac{5}{10}[/tex].
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 [tex]p_{R}=\frac{3}{10}[/tex].
The probability mass function of a Binomial distribution is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3....[/tex]
Compute the probability of selecting less than 2 red balls in 3 draws as follows:
P (R < 2) = P (R = 0) + P (R = 1)
[tex]={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[/tex]
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.