Question

A bag contains 5 black & 6 white balls. If two

balls are drawn together at random, then the

probability that these being of different colour​

Answer 1

Given :-----

• Black balls = 5
• white balls = 6
• it has been said that 2 balls are drawn together at random ..

To Find :------

• Probability of both balls are of different color ...

we can solve it by 2 methods ... since balls are drawn randomly , we will use combination formula ...

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$\Large\bold\star\underline{\underline\textbf{Solution(1)}}$

we know that, combination Formula = nCr = n!/(n-r)!*r!

Black balls = 5 and white balls are = 6 ,

Total balls = 11 = n

2 balls choose at random , so r = 2 ..

putting in formula we get,,,

→ 11!/(11-2)!*2!

→ 11×10×9!/9!*2!

→ 11×10/2×1

→ 55 = Total Number of ways of choosing 2 balls ...

Now,

Probability of both balls are of black color =

→ 5!/(5-2)!*2!

→ 5×4×3!/3!*2!

→ 5×4/2

→ 10

Similarly ,,,

Probability of both balls of white color =

→ 6!/(6-2)!*2!

→ 6×5×4!/4!*2!

→ 6×5/2

→ 15

Total probability of black and white balls = 10+15 = 25 .

Hence,

Probability that both balls are of same color = Favourable outcome /Total no. of outcome

→ P = 25/55 = 5/11 ...

Now,

we have to find probability of both balls are of different color , that means we will subtract same color with total 1 outcome ..

Hence,

Required probability = 1 - 5/11 = 6/11 ...

$\red{\large\boxed{\bold{ \frac{6}{11} }}}$

is our Required Answer.....

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$\Large\bold\star\underline{\underline\textbf{Solution(2)}}$

Lets try to do it in easy way ....

For probability that both balls are of different colour you have to choose two different colour of balls.

Let first ball choosen be red and other be white.

So,,,, Required Probability is : ---------

→ (5c1×6c1)/11c2

→ 5×6/55

$\pink{\large\boxed{\boxed{\bold{ \frac{6}{11} }}}}$

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$\large\underline\textbf{Hope it Helps You.}$