Question

A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is

Answer 1

Given   : A box contains 4 red balls and 6 black balls.

Three balls are selected randomly from the box one after another, without replacement.

To Find : The probability that the selected set contains one red ball and two black balls is

Solution:

Red = 4

Black = 6

Total = 10

3 out of 10 can be selected in ¹⁰C₃  = 120    

set contains one red ball and two black balls

Number of ways

⁴C₁*⁶C₂

=  4 * 15

= 60

probability that the selected set contains one red ball and two black balls is = 60/120

= 1/2

= 0.5

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Answer 2

Question= A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is

To find:

The probability that the selected set contains one red ball and two black balls is

Solution:

Correct answer: 1/2

So here in total we have 10 balls , which is a finite population and out of which we have 4 red and 6 black balls..Now according to the question , favorable event is the one in which we have one red balls and two black balls..

Total number of events = Number of ways in which we can pick 3 out of 10 balls = 10C3

Now number of favorable events = Number of ways 1 red ball can be picked out of 4 red balls * Number of ways 2 black balls can be picked from 6 balls = 4C1 * 6C2

Hence required probability = ( 4C1 * 6C2 ) / 10C3

= ( 4 * 15 * 6 ) / ( 10 * 9 * 8 )

= 1 / 2