A box contains 4 red pens, 5 blue pens, and 9 green pens. Two pens are drawn out of the box without replacement. What is the probability of drawing a red pen on the first draw and the same color pen on the second draw?
The probability of a red pen on the first draw is 4/18.
For the second draw, there are 3 red pens left out of 17 pens total.
The probability of both happening is obtained by multiplying the two probabilities:
P(RR)=(4/18)*(3/17)=2/51
To find the probability of drawing a red pen on the first draw and the same color pen on the second draw, we need to find two separate probabilities and multiply them together.
First, let's find the probability of drawing a red pen on the first draw. The total number of pens in the box is 4 + 5 + 9 = 18. Since there are 4 red pens, the probability of selecting a red pen on the first draw is 4/18.
After drawing a pen on the first draw, there will be one less pen in the box. Therefore, for the second draw, there will be a different total number of pens in the box, depending on the color of pen drawn on the first try.
If a red pen was drawn on the first try, there will be 3 red pens remaining out of a total of 17 pens. So, the probability of selecting a red pen on the second draw, given that a red pen was drawn on the first draw, is 3/17.
Finally, to find the overall probability of drawing a red pen on the first draw and the same color pen on the second draw, we multiply these two probabilities together:
(4/18) * (3/17) = 12/306 ≈ 0.0392
Therefore, the probability is approximately 0.0392, or 3.92%.