00:01
So the distribution of each color of m &m found in a sample bag of 280 m &ms is recorded.
00:09
And we want to reach into the bag and randomly select one m &m and calculate three different probabilities.
00:18
And our first probability is going to be the probability of red.
00:29
So we think of probability as favorable over possible.
00:33
And the favorable red m &ms was 39 out of a possible 280 m &ms in that bag.
00:46
So therefore your probability would be 0 .139 -285 -7143.
00:57
But your directions were very specific to round to three decimal places.
01:02
So we would say the probability is 0 .13.
01:08
For part b, we want to determine the probability that when we select one m &m, it is yellow.
01:17
So favorable over possible, there are 55 favorable yellow m &ms out of a possible 280 in the bag.
01:27
As a decimal, that would equate to .196, 4, 28, 5714.
01:37
So rounding to three decimal places, we would say the probability is 0 .1964.
01:45
For letter c, we want to determine the probability of reaching in the bag and selecting one blue m &m.
01:54
So favorable over possible would be 27 favorable blue m &ms out of a possible 280 total m &ms, giving you a probability of 0 .090.
02:09
964 -2 -85714, 4 .0964.
02:19
Now, as we move on to part two of this, this time we're going to reach into the bag and randomly select three m &ms, and we're going to talk about probability with replacement and probability without replacement.
02:34
So in part four, or part d i'm calling, the probability that the first m &m is red, followed by the second m &m being yellow, and the third m &m is, and the page was cut off, i'm assuming it's supposed to be blue based on the first three problems.
03:00
So as long as we are doing this with replacement, what we'll do is we'll calculate each probability.
03:09
Separately and then just multiply them together.
03:13
So we'll take the probability of red and multiply it by the probability of yellow and multiply it by the probability of blue.
03:24
So that means we're taking 39 out of 280 and multiplying it by 55 out of 280 and multiplying it by 25 out of 280 and multiplying it by 27 out of 280.
03:39
And that is going to give you a probability of 0 .006, 382, 56, and to four decimal places, we'd say 0 .0026.
04:01
For part e, we want to do the same probability of the first one being red, the second one being yellow, and the third one being blue, but this time we want without replacement.
04:32
So when we're doing this without replacement, we're still going to find the probability of red, but then we're going to multiply it by the probability of yellow given that a red was already drawn, and we'll multiply that by the probability of blue, given that we've already had a red and a yellow.
05:00
So the probability of a red is 39 out of the 280.
05:07
Now, once we get a red out of there, there's no longer 280.
05:13
There's now 38 reds in the bag, and there's 279 total in the bag.
05:20
So when it comes time to find our probability of yellow, there are 55 favorable yellow marbles, but only 279 in the bag.
05:32
And then once we get a yellow marble, there's now 54 yellow marbles in the bag and 278 total in the bag...