Question

Box a contains 5 red and 4 blue balls, box b contains 2 red and 5 blue balls. A ball is drawn at random from each box. Find the probability that one is red and the other is blue.

Answer 1

Answer:

total balls in box A=5+4

=9

probability (drawn ball is red)=no. of favourable outcome/total outcomes

=5/9

total balls in box b=2+5=7

probability(drawn ball is red)=no of favourable outcome/total no of outcomes

=5/7

Answer 2

Answer:

the required probability is $\frac{11}{21}$

Step-by-step explanation:

as per the given question,

we are given 2 boxes A and B in which

A contains 5 Red and 4 blue balls

then there is a total of 9 balls in Box A

and in box B there are 2 red and 5 blue balls

so box B contains 7 balls.

according to question , we need to find the probability that if 1 ball is drawn at random from each box, the balls is one red and other is blue

so there are 2 cases,

that we draw a red ball from A and a blue from B or we can get a red from B and blue from A

$P(red from A)=\frac{^5C_1}{9}$

$P(blue from B)= \frac{^5C_1}{7}$

$P( red from B) =\frac{^2C_1}{7}$

$P(blue from A)= \frac{^4C_1}{9}$

so needed probability is ,

$\frac{^5C_1}{9}*\frac{^5C_1}{7} + \frac{^4C_1}{9}*\frac{^2C_1}{7}$

on solving ,

$\frac{5}{9}*\frac{5}{7}+\frac{4}{9} *\frac{2}{7} \\= \frac{25}{63}+\frac{8}{63} \\=\frac{33}{63} \\=\frac{11}{21}$

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