Question

A box contains 4 black, 3red and 6 green marbles. 2 marbles are drawn from the box at random. what is the probability that both the marble are out of the same colour?
a) 12/74
b) 24/78
c) 13/78
d) None of those​

Answer 1

Answer:

the answer is 1/13. so none of these

Answer 2

The probability that both the marble are out of the same colour = 24/78

Given :

• A box contains 4 black, 3 red and 6 green marbles.

• 2 marbles are drawn from the box at random

To find :

The probability that both the marble are out of the same colour

a) 12/74

b) 24/78

c) 13/78

d) None of those

Solution :

Step 1 of 3 :

Calculate total number of possible outcomes

The box contains 4 black, 3 red and 6 green marbles.

Total number of marbles = 4 + 3 + 6 = 13

2 marbles are drawn from the box at random.

So total number of possible outcomes

$\displaystyle \sf{ } = {}^{13} C_2$

$\displaystyle \sf{ = \frac{13!}{2!(13 - 2)!} }$

$\displaystyle \sf{ = \frac{13!}{2! \: 11!} }$

$\displaystyle \sf{ = \frac{13 \times 12 \times 11!}{2! \: 11!} }$

$\displaystyle \sf{ = \frac{13 \times 12 }{2! } }$

$\displaystyle \sf{ = \frac{13 \times 12 }{2 } }$

$\displaystyle \sf{ = 78}$

Step 2 of 3 :

Calculate total number of possible outcomes for the event

Let A be the event that both the marble are out of the same colour

Then three cases arises :

Case : 1

Both marbles are black

Total number of black marbles = 4

The number of ways 2 marbles can be taken from 4 marbles

$\displaystyle \sf{ } = {}^{4} C_2$

$\displaystyle \sf{ = \frac{4!}{2!(4 - 2)!} }$

$\displaystyle \sf{ = \frac{4!}{2! \: 2!} }$

$\displaystyle \sf{ = \frac{4 \times 3 \times 2!}{2! \: 2!} }$

$\displaystyle \sf{ = \frac{4 \times 3 }{2! } }$

$\displaystyle \sf{ = \frac{12 }{2 } }$

= 6

Case 2

Both marbles are red

Total number of red marbles = 3

The number of ways 2 marbles can be taken from 3 marbles

$\displaystyle \sf{ } = {}^{3} C_2$

$\displaystyle \sf{ = \frac{3!}{2!(3 - 2)!} }$

$\displaystyle \sf{ = \frac{3!}{2! \: 1!} }$

= 3

Case : 3

Both marbles are green

Total number of green marbles = 6

The number of ways 2 marbles can be taken from 6 marbles

$\displaystyle \sf{ } = {}^{6} C_2$

$\displaystyle \sf{ = \frac{6!}{2!(6 - 2)!} }$

$\displaystyle \sf{ = \frac{6!}{2! \: 4!} }$

$\displaystyle \sf{ = \frac{6 \times 5 \times 4!}{2! \: 4!} }$

$\displaystyle \sf{ = \frac{6 \times 5}{2! } }$

$\displaystyle \sf{ = \frac{30 }{2 } }$

= 15

So total number of possible outcomes for the event A

= 6 + 3 + 15

= 24

Step 3 of 3 :

Find the probability

Hence the required probability

= P(A)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: A }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{24}{78} }$

$\displaystyle \sf{ = \frac{12}{39} }$

Hence the correct option is b) 24/78

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