Many has two boxes one contains 10 colored pencils 4 of witch are red and 6 which are green the other box onther box contains 7 markers and which 4 are red and 3 are blue what's the probability of of many choosing green green pencil and red marker
Pr(grpencil+redmarker)=6/10 * 4/7
=24/70=12/35 about one in three.
To find the probability of Many choosing a green pencil from the first box and a red marker from the second box, we need to take into consideration the number of favorable outcomes and the number of total outcomes.
First, let's determine the number of favorable outcomes. Many needs to choose a green pencil from the first box, which contains 6 green pencils out of a total of 10 colored pencils. So, the probability of Many choosing a green pencil from the first box is 6/10.
Next, Many needs to choose a red marker from the second box, which contains 4 red markers out of a total of 7 markers. Hence, the probability of Many picking a red marker from the second box is 4/7.
To find the probability of both events happening, we multiply the two probabilities together. Thus, the probability of Many choosing a green pencil from the first box and a red marker from the second box is (6/10) * (4/7).
Calculating it gives us (6/10) * (4/7) = 24/70.
Simplifying the fraction, we get 6/17.
Therefore, the probability of Many choosing a green pencil from the first box and a red marker from the second box is 6/17.