Question

One bag contains 4 white balls and 2 black balls. Another bag contains 3 white balls and 5 black balls. Ifone
ball is drawn from each bag, what is the probability that one ball is white and another ball is black?​

Answer 1

Answer:

Step-by-step explanation:

Suppose that you have the bag. Of course, segregation is not justified, so we ain't use the separation technique for that.

The total balls in the bag will be the whole lot; ie the bag with 4 white balls and 2 black balls as one unit  (1) and 3 white balls and 5 black balls as another (2) .

Now, considering the cases with black balls :-

Black balls of bag 1 = $\sf{\dfrac{2}{4 + 2} = \dfrac{2}{6} = \dfrac{1}{3}}$

Black balls of bag 2 = $\sf{\dfrac{5}{8}}$

Now, case 2 with white balls :-

Black balls of bag 1 = $\sf{\dfrac{4}{4 + 2} = \dfrac{4}{6} = \dfrac{2}{3}}$

Black balls of bag 2 = $\sf{\dfrac{3}{5+3} = \dfrac{3}{8}}$

But, what we want is one ball is white and another ball is black, so we will take each cases and cross it with each other.

So, there will be one black and white ball case from each bags (bag 1 and 2)

$\sf{\dfrac{5}{8} \times \dfrac{2}{3}}$  = $\sf{\dfrac{10}{24}}$

$\sf{\dfrac{1}{3} \times \dfrac{3}{8} = \dfrac{3}{64}}$

Now, we will add it because we want it as whole!

[Using it as nominal form can be used as it's shortcut]

$\sf{\dfrac{4}{6} \times \dfrac{5}{8} = \dfrac{20}{48}}$

$\sf{\dfrac{2}{6} \times \dfrac{3}{8} = \dfrac{6}{48}}$

$\sf{\dfrac{20 + 6}{48}}$

$\sf{\dfrac{26}{48}}$ is the required answer. Simplifying, we get

$\sf{\dfrac{13}{24}}$ as the answer.

Answer 2

Given :-

One bag contains 4 white balls and 2 black balls. Another bag contains 3 white balls and 5 black balls. If one ball is drawn from each bag,

To Find :-

what is the probability that one ball is white and another ball is black?​

Solution :-

At first

$\sf Black \: balls (1) = \dfrac{2}{6} = \dfrac{1}{3}$

$\sf Black \: Balls(2) = \dfrac{5}{6}$

Now,

$\sf White \: balls (1) = \dfrac{4}{6} = \dfrac{2}{3}$

$\sf White \: balls (2) = \dfrac{3}{8}$

So,

For probability

$\sf White_{(1)} and \: Black_{(2)} = \dfrac{4}{6} \times \dfrac{5}{8}$

$\sf White_{(1)} and \: Black_{(2)}= \dfrac{1}{6} \times \dfrac{5}{2}$

$\sf White_{(1)} and \: Black_{(2)} = \dfrac{5}{12}$

$\sf White_{(2)} and \: Black_{(1)} = \dfrac{2}{6} \times \dfrac{3}{8}$

$\sf White_{(2)} and \: Black_{(1)} = \dfrac{1}{2 \times 4}$

$\sf White_{(2)} and \: Black_{(1)} = \dfrac{1}{8}$

Probability = 5/12 + 1/8

Probability = 20 + 6/48

Probability = 26/48 = 13/24