S2 - exam question??

Binomial expansion assumes statistical independence so P(A and B and C)=P(A)P(B)P(C) But there are 3 boxes. If you use the combinations that can arise

Let P(D) be the probability when one egg is double yolk and thus D is this event.
D' is the event not occurring.

We want 2 with one double yolks and one without. So DDD', this can have different combinations right. We need to times the combinations by the probability of getting DDD'.



Posted from TSR Mobile

So in that sense couldnt I just do three choose two multiplied by (0.3412)^2 and that multiplied by 0.6588?

But we want combinations and not to choose.


Posted from TSR Mobile

Sorry I read the question wrong for part c. The reason for the choosing now is that 3 boxes have been chosen at random.
Let a = P(D) and b = P(D'. Since we chose 3 at random we obtain a new expansion (a+b)^3 and we are interested in the a^2 b term so it is 3 choose 2


Posted from TSR Mobile

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Just read the above exchange and I have no clue what's going on. You're all overcomplicating this. This is a simple case of using the binomial distribution. Let $X$ be the random variable "# of boxes containing exactly 1 egg with a double yolk" then you have 3 boxes, so that's $n=3$ and the probability of a double yolk is $p =0.3412$.

i.e: you have $X \sim B(3, 0.412)$ and you want to find "2 boxes containing exactly 1 egg with a double yolk" so you want $\mathbb{P}(X =2)$.

Now look in your formula booklet and you should find that $\mathbb{P}(X=2) = {3 \choose 2}(0.3412)^2 (1-0.3412)^1$.

This is basically the same thing as you did except you forgot the missing factor of ${3 \choose 2}$.

That's because you did 3 boxes: probability box 1 contains * probability box 2 contains * probability box 3 doesn't contain.

But really, there are three boxes: so you need to add all three probabilities up:

probability box 1 contains * probability box 2 contains * probability box 3 doesn't contain
probability box 1 contains * probability box 3 contains * probability box 2 doesn't contain
probability box 2 contains * probability box 3 contains * probability box 1 doesn't contain

The binomial distribution was invented precisely to avoid listing all these options down which gets tiresome when you do things with more than just 3 boxes, so use it. You'll get used to spotting it once you do more practice.

So sum up, you have a binomial distribution with $X\sim B(3, 0.3142)$ and you want $\mathbb{P}(X=2)$.

Wow. Now that makes a lot more sense! Yeah I wasn't really sure during the above exchange either, this is much more logical though actually, thank you once more Zain

No problem. Sorry for the later reply, was watching a movie. :-)

Don't worry - I'm lucky to have received help anyway haha! Anything good?

@Zacken @aymanzayedmannan and other maths wizards to the rescue! <3

No problem. Sorry for the late reply, was watching a movie. :-)

Go do STEP!!!