00:01
All right, so we're told that in a bag of 80 m &ms, we have 10 red, 11 orange, 18 blue, 10 green, 16 yellow, and 15 brown.
00:20
Okay? and once again, that's out of 80.
00:23
And they're equally mixed up.
00:25
So a asks, what's the probability of picking a red one? okay.
00:32
So we know that there's 10 reds out of the 80, right? so the probability for that to 3 decimal places, that's going to be 1 8th, right? so that's 0 .125.
00:44
For b, if we select 1 at random, what's the probability of not green? okay.
00:52
Well, we know that there's 10 that are green, right? which means that the other 70 out of the 80 are not.
01:01
Right so if i take 70 divided by 80 i get 0 .875 so that's for b right for c if we select one at random what is the probability that is red or yellow so we know that we have 10 red and 16 yellow so that's a total of 26 that are red or yellow right so that's going to be 26 out of the 80 which is 0 .325 okay so that's the probability for c.
01:34
For d, we've got, if we select one at random, then put them all back and mix them up again, then what's the probability that we get the both and second ones, both green? okay, so we know that for each trial here, the probability of getting a green is 10 out of 80, right? but we're doing that twice in a row, which means we have to do 10 out of 80 times 10 out of 80, or an other words, 1 eighth or 10 out of 80 squared...