Question

four red, 8 blue, and 5 green balls are randomly arranged in a line. (a) what is the probability that the rst 5 balls are blue? (b) what is the probability that none of the rst 5 balls are blue? (c) what is the probability that the nal 3 balls are dierently colored. (d) what is the probability that all the red balls are together?

Answer 1

Answer:

We have been given the total number of balls =4+8+5=17

(a) $Probability=\frac{\text{Favourable outcomes}}{\text{Total number of outcomes}}$

$=\frac{8}{17}$

(b)$\text{Balls are not blue}=1-\text{balls are blue}$

$\text{Balls are not blue}=1-\frac{8}{17}$

$\frac{9}{17}$

(c) Probability of all three balls are differently colored $\frac{^4C_1\cdot ^8C_1\cdot ^5C_1}{^{17}C_3}$

Using:$^nC_r=\frac{n!}{r!\cdot (n-r)!}$

On simplification we get:

$\frac{4}{17}$

(d) Probability that all red balls are together $\frac{^4C_4}{^{17}C_4}$

On simplification we get:

$0.000017$