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A bag contains red marbles and blue marbles. If two marbles are chosen at random without replacement, the probability that both marbles are red is , and the probability that both marbles are blue is also . How many marbles are in the bag?

Sagot :

Answer:

So, a + b = 3 + 3 = 6

Hence,

There will be 6 marbles in the bag, 3 red and 3 blue.

Step-by-step explanation:

In this question probabilities are not given which should be given in order to calculate the number of marbles in the bag.

So, let's assume the probability that both marbles are red = 0.5

And probability that both marbles are blue = 0.5

Total Probability = 1

Furthermore, Let's suppose the bag contains

a number of red marbles and b number of blue marbles.

This problem includes the concepts of combination.

So,

[tex]\frac{_{2} C^{a} }{_{2} C^{a+b} }[/tex] = 0.5      Equation 1

[tex]\frac{_{2} C^{b} }{_{2} C^{a+b} }[/tex] = 0.5      Equation 2

Dividing Equation 1 and equation 2  we will get

[tex]\frac{_{2} C^{a} }{_{2} C^{b} }[/tex] = 1

It means, we can write:

[tex]_{2} C^{a} = _{2}C^{b}[/tex]

Hence, we can conclude that:

a = b

Now, from equation 1, we can write:

[tex]\frac{_{2} C^{a} }{_{2} C^{2a} }[/tex] = 0.5

Breaking down the combinations, we will get:

[tex]\frac{\frac{a!}{2! (a-2)!} }{\frac{2a!}{2! (2a - 2)!} }[/tex] = 0.5

[tex]\frac{a ( a-1)}{2a (2a-1)} = \frac{1}{5}[/tex]

(a-1)5 = 4a - 2

5a - 5 = 4a - 2

5a -4a = 5-2

a = 3

As, we know, a = b

So, a + b = 3 + 3 = 6

Hence,

There will be 6 marbles in the bag, 3 red and 3 blue.

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