1.

A box contains 8 balls, 2 red, 3 blue , 1 green and 2 white. Four balls are selected randomly. What is the probability of having?1) 2 red and 2 blue?2) Different balls?3) At least 2 blue balls?​

Answer»

Answer:

First of all we have 10 balls of which 2 are red, 3 are black and 4 are blue.

If we can think carefully, the question is given that, when 3 balls are drawn, the balls are of DIFFERENT COLORS. There are 3 colors. So definitely one BALL should be drawn from each of the color.

So the number of ways of HAPPENING the above case is 4C1*3C1*2C1.

Let it be T(E).

The number of ways of drawing three balls from 9 balls is given as 9C3. Let this be S(E).

Let the probability of happening the given condition be P(E).

Then according to definition of probability, we calculate it as,

P(E)=T(E)/S(E).

P(E)=4*3*2/9C3.

=24/84.

=2/7.


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