Aptitude 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 $

$= \textbf{21/40}$

**Syayli**

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**Naveen**

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

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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**

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21/40 is the right answer.

**Avinash**

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

Check it dear..

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

Somehow miscalculated it

**Rohit**

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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.

**???? ???**

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

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

can someone solve it using permutation combination method plz