A box contains 9 chips of which 4 are blue and remaining red...

A box contains 9 chips, of which 4 are blue and remaining red. We selected 3 chips at random. Let X denote the number of blue chips in our selection. (a) X follows Hyper-geometric distribution. (b) P(X = 3) = .024 (c) P(X = 0)= .212 (d) P(X=1)= .505 (e) P(X≤ 1)= .717 (f) P(X > 1)= .065

Transcript

00:01 Okay, so we have a box with 12 balls.

00:04 There's 7 blue, 4 red, and 1 green.

00:12 So that's 12 total.

00:15 And we're going to choose 2 without replacement.

00:18 Okay? so we want the different probabilities.

00:26 Okay? so for the first one, we want the probability that both are blue.

00:31 Okay? so you just have to think of this as two separate things, right? you could have the...

00:43 So if the first ball is blue, there's a 7 out of 12 chance of that.

00:47 And then now, for the next ball, there's only 6 blue balls left out of the 11 that are left.

00:55 So the probability that both are blue is going to be 7 over 12 times 6 over 11, which is .318 if i'm rounding to 3 decimals.

01:10 Okay? for b, we want the probability that both are green.

01:17 Well, the probability that the first ball is green is 1 out of 12.

01:22 The probability that the second ball is green is then 0 out of 12 because there's only 1 green ball.

01:27 Right? so that makes it 0.

01:28 The probability of that happening is 0.

01:30 Because there's only 1 green ball in the first place.

01:33 For c, for the probability that both are red, you've got a 4 out of 12 chance that the first one is red, and then a 3 out of 12 chance that the second one is red then.

01:45 Right? so that is 4 twelfths times 3 twelfths, which comes out to .083.

01:56 For d, the probability that the first is blue and the second is green...

02:05 Well, the probability that the first one is blue, we know, is 7 out of 12...

Question

A box contains 9 chips, of which 4 are blue and remaining red. We selected 3 chips at random. Let X denote the number of blue chips in our selection. (a) X follows Hyper-geometric distribution. (b) P(X = 3) = .024 (c) P(X = 0)= .212 (d) P(X=1)= .505 (e) P(X≤ 1)= .717 (f) P(X > 1)= .065

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