# Easy Probability Solved QuestionAptitude Discussion

 Q. A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag. Find the probability that one ball is red and one is green.
 ✖ A. 19/20 ✖ B. 17/20 ✖ C. 8/10 ✔ D. 21/40

Solution:
Option(D) is correct

Let $A$ be the event that ball selected from the first bag is red and ball selected from second bag is green.

Let $B$ be the event that ball selected from the first bag is green and ball selected from second bag is red.

$P(A)= (5/8) ×(6/10)$

$= 3/8$ and

$P(B)= (3/8) ×(4/10)$

$= 3/20$

Hence, required probability,

$= P(A)+P(B)$

$= 3/8+ 3/20$

$= \textbf{21/40}$

## (9) Comment(s)

Syayli
()

can someone solve it using permutation combination method plz

Naveen
()

$P(A)=\left(\dfrac{5}{8}\right) \times \left(\dfrac{6}{10} \right)= \dfrac{3}{8}$ and

$P(B)=\left(\dfrac{3}{8}\right) \times \left(\dfrac{4}{10} \right)= \dfrac{3}{20}$

Hence required probability $=P(A)+P(B)$

$=\dfrac{3}{8}+\dfrac{3}{20}$

$= \dfrac{21}{40}$

Naveen
()

Since $P(A)=\dfrac{3}{8}$ and $P(B)=\dfrac{3}{20}$

$P(A) + P(B) = \dfrac{3}{8}+\dfrac{3}{20}$

$\Rightarrow P(A) + P(B) =\dfrac{(5*3)+(2*3)}{40}$

$=\dfrac{(15+6)}{40}$

$=\dfrac{21}{40}$

Uttiyan
()

Avinash
()

@ Rumila how have u got $\dfrac{3}{8}+\dfrac{3}{20}$ as $\dfrac{9}{10}$???

Check it dear..

Rumila
()

I got that wrong, it should have been $\dfrac{3}{4}+\dfrac{3}{20}=\dfrac{9}{10}$

Somehow miscalculated it

Rohit
()

I guess P(A) is calculated wrong. Its value should be $\dfrac{3}{8}$ and not $\dfrac{3}{4}$. Please make the necessary corrections.

P.S. You people are doing great job, keep it up.

???? ???
()

$P(A)= \dfrac{3}{8}$ AND $P(A)+P(B)=\dfrac{21}{40}$

Rumila
()

Since $P(A)=\dfrac{3}{8}$ and $P(B)=\dfrac{3}{20}$

$P(A) + P(B) = \dfrac{3}{8}+\dfrac{3}{20}$

$\Rightarrow P(A) + P(B) = \dfrac{(5*3)+3}{20}$

$\Rightarrow P(A) + P(B) =\dfrac{18}{20} = \dfrac{9}{10}$

So, I guess the answer is correct