Question

A box having 5 black and 3 brown flags. another box having 4 black and 6 brown flags. if one flag is drawn from each box. find the probability that both flags are of different color.

Answer 1

1/7and1/10are the probability of getting the flag

Answer 2

Answer:

$\text{The required probability is : }\frac{21}{40}$

Step-by-step explanation:

Probability (1st black and 2nd brown) = P(black from the first box) × P(brown from the second box)

$=\frac{5}{8}\times \frac{6}{10}=\frac{30}{80}=\frac{3}{8}$

Probability (1st brown and 2nd black) = P(brown from the first box) × P(black from the second box)

$=\frac{3}{8}\times \frac{4}{10}=\frac{12}{80}=\frac{3}{8}$

⇒ P(different color) = Probability (1st black and 2nd brown) + Probability (1st brown and 2nd black)  

$\implies \text{P (different color) = }\frac{3}{8}+\frac{12}{80}=\frac{42}{80}=\frac{21}{40}$