Question

There are 2 blue balls, 4 black balls and 3 white balls in a box. If two balls are taken out from the box one after the another. Then what is the probability that there is no white ball in these 2?
a) 3/12 b) 5/12 c) 7/12 d) 9/12

Answer with proper explanation.​

Answer 1

Answer:

$\boxed{\bf \: b) \: \: \: \dfrac{5}{12} \: }$

Step-by-step explanation:

Given that, there are 2 blue balls, 4 black balls and 3 white balls in a box.

So, Total number of balls in a box = 2 + 4 + 3 = 9

Now, we have to take out two balls one after the other, so that there is no white ball in these two balls.

Let assume the following events:

E₁ : The ball drawn is not of white color in first draw

E₂ : The ball drawn is not of white color in second draw.

So, required probability of not getting a white ball is

$\sf \: = \: P(E_1 \: \cap \: E_2)$

$\sf \: = \: P(E_1) \times P(E_2\: \mid \: E_1)$

$\sf \: = \: \dfrac{6}{9} \times \dfrac{5}{8}$

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

$\sf \: = \: \dfrac{1}{3} \times \dfrac{5}{4}$

$\sf \: = \: \dfrac{5}{12}$

Hence,

$\implies\boxed{\sf \: Probability \: of \: not \: getting \: white \: balls \: in \: 2 \: draws \: is \: \dfrac{5}{12}}$

Answer 2

Step-by-step explanation:

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