Question

One bag contains 8 blue balls and 6 Green balls; another bag contains 7 blue balls
and 5 green balls
If one ball is drawn from each bag, determine the probability that both are blue?
*
O 1/2
O 1/3
O 1/4
h
O 1/5​

Answer 1

Answer:

answer is 1/4

Step-by-step explanation:

total no of balls n(s) =8+6+7+5=24

no of blue ballsn(e)=6

P(E)=n(E)/n(S)=>6/24=1/4

MARK IT AS BRAINLIST ANSWER

Answer 2

Answer:

The probability that both balls are blue is equal to 1/3.

Therefore, the option (2) is correct.

Step-by-step explanation:

Given, the number of blue balls in 1st bag = 8

the number of green balls in 1st bag = 6

The total number of balls in 1st bag = 8 + 6 =14

Probability = number of favorable events/ total number of events

Probability of drawn blue ball from first bag :

$P(E_{1})=\frac{^{8}C_{1}}{^{14}C_{1}}$

Here, $^{n}C_{r}=\frac{n!}{r!(n-r)!}$

$P(E_{1})=\frac{8!}{1!*(8-1)!} *\frac{1!(14-1)!}{14!}$

$P(E_{1})=\frac{8!}{7!} *\frac{13!}{14!}$

$P(E_{1})=\frac{8}{14}$

Number of blue balls in 2nd bag = 7

the number of green balls in 2nd bag = 5

The total number of balls in 2nd bag = 7 + 5 =12

Probability of drawn blue ball from second bag :

$P(E_{2})=\frac{^{7}C_{1}}{^{12}C_{1}}$

Here, $^{n}C_{r}=\frac{n!}{r!(n-r)!}$

$P(E_{2})=\frac{7!}{1!*(7-1)!} *\frac{1!(12-1)!}{12!}$

$P(E_{2})=\frac{7!}{6!} *\frac{11!}{12!}$

$P(E_{2})=\frac{7}{12}$

The selection of both balls will be independent from each other, so we multiple the both probabilities.

Probability that both balls are blue $= P(E_{1})*P(E_{2})$

$=(\frac{8}{14}) *(\frac{7}{12})$

$=\frac{1}{3}$

Therefore, the probability of drawn ball from each bag is blue will be equal to 1/3.