# Answer to Question #370 in Statistics and Probability for Alex Landon

Question #370

There are 10 red and 6 blue buttons in the box. What is the probability that two randomly taken buttons are of the same color?

Expert's answer

Extraction of two buttons can be represented as a sequential extraction of one button, and then another. The probability of taking the blue button on the first pulling is: P(blue, 1) = 6/(10+6) = 6/16. If we have got the blue button at the first pulling, then there are 10 red and 5 blue buttons left in the box, which means that the probability to pull the second blue button is: P(blue, 2) = (6-1)/(10+6-1) = 5/15.

The results of the two described extractions are not dependent on each other, hence the probability of getting two blue buttons (a blue button at the first extraction and a blue button at the second extension) is: P(blue) = P(blue, 1)*P(blue, 2) = (6/16)*(5/15) = 30/(16*15) = 1/8.

Similarly for the red buttons:

P(red, 1) = 10/(10+6) = 10/16;

P(red, 2) = (10-1)/(10+6-1) = 9/15;

P(red) = P(red, 1)*P(red, 2) = (10/16)*(9/15) = 90/(16*15) = 3/8.

Getting two buttons of the same color is equivalent to getting two red or two blue buttons. As these events are independent, the probability of extracting two buttons of the same color is:

P(blue or red) = P(blue) + P(red) = 1/8 + 3/8 = 4/8 = 1/2.

The results of the two described extractions are not dependent on each other, hence the probability of getting two blue buttons (a blue button at the first extraction and a blue button at the second extension) is: P(blue) = P(blue, 1)*P(blue, 2) = (6/16)*(5/15) = 30/(16*15) = 1/8.

Similarly for the red buttons:

P(red, 1) = 10/(10+6) = 10/16;

P(red, 2) = (10-1)/(10+6-1) = 9/15;

P(red) = P(red, 1)*P(red, 2) = (10/16)*(9/15) = 90/(16*15) = 3/8.

Getting two buttons of the same color is equivalent to getting two red or two blue buttons. As these events are independent, the probability of extracting two buttons of the same color is:

P(blue or red) = P(blue) + P(red) = 1/8 + 3/8 = 4/8 = 1/2.

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