Probability Question 389
Question: A bag contains an assortment of blue and red balls. If two balls are drawn at random, the probability of drawing two red balls is five times the probability of drawing two blue balls furthermore, the probability of drawing one ball of each color is six times the probability of drawing two blue balls. The number of red and blue balls I the bag is
Options:
A) 6, 3
B) 3, 6
C) 2, 7
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let the number of red and blue balls be r and b, respectively.
Then, the probability of drawing two red balls $ =p_1=\frac{^{r}C_2}{^{r+b}C_2}=\frac{r(r-1)}{(r+b)(r+b-1)} $
The probability of drawing two blue balls is $ p_2=\frac{^{b}C_2}{^{r+b}C_2}=\frac{b(b-1)}{(r+b)(r+b-1)} $
The probability of drawing one red ad one blue ball $ =p_3=\frac{^{r}C_1^{b}C_1}{^{r+b}C_2}=\frac{2br}{(r+b)(r+b-1)} $
By hypothesis, $ p_1=5p_2 $ and $ p_3=6p_2. $
Therefore, $ r(r-1)=5b(b-1) $ and $ 2br=6b(b-1) $
$ \Rightarrow r=6,b=3. $
BETA
