Probability, Conditional probability, Normal distribution

Question

I have 3 independently random variables A, B, C, all of N (10,1). What is the probability that the largest of A, B, C is greater than 13 ?

My solution

$P[A>13 | A-B>0 \wedge B-C>0]$ $= \frac{1-P[A<13]}{(1-P[A-B<0])(1-P[A-C<0])}$$= \frac{1-\Phi (13-10)}{(1-\Phi (0))^{2}}=\frac{1-0.99865}{\frac{1}{4}}$

It is correct ? Thanks :)

Answer 1

Since

$$\Pr[\max(A,B,C)\leq z] = \Pr[A\leq z]\Pr[B\leq z]\Pr[C\leq z] = F^3(z),$$

we have

$$\Pr[\max(A,B,C)>13)] = 1 - \Pr[\max(A,B,C)\leq z] = 1 - F^3(13).$$

Answer 2

Let $X_1, ..., X_n \sim \mbox{iid } F$ and define $X^* \equiv \max\{X_1, ..., X_n\}$. Then, $$\mathbb{P}(X^*\leq x) = \mathbb{P}(X_1 \leq x \cap \cdots \cap X_n \leq x) = \mathbb{P}(X_1 \leq x)\cdots \mathbb{P}(X_n \leq x) = F^n(x)$$ It follows that $$\mathbb{P}(X^* > x) = 1 - \mathbb{P}(X^* \leq x) = 1 - F^n(x) $$ In your example $n=3$, and $F(x) = \Phi\left(x - 10\right)$, so the probability is $$1 - \Phi^3(13 - 10) = 1 - \Phi^3(3) \approx 0.004044$$

Answer 3

The largest of these is greater than $13$ provided that at least one of them is greater than $13$.

So the probability is $$1-p(\text{all 3 less than 13})$$ $$=1-(0.9986501019...)^3=0.00404423...$$

Answer 4

The correct answers have been provided by others so this post will be about what the OP did wrong. Hopefully the OP is a student who wants to learn from this...

Rename $E \equiv (A > 13); F \equiv (A>B); G \equiv (B>C)$ then the OP's first equation is

$$ P(E | F \wedge G) \overset{???}{=} \frac{P(E)}{P(F) P(G)} $$

but that is incorrect. The correct formula is:

$$ P(E | F \wedge G) = \frac{P(E \wedge F \wedge G)}{P(F \wedge G)}$$

• The OP numerator is clearly wrong. The correct numerator is also hard to evaluate (unless you use the methods in the other posted answers).

• The OP denominator is also wrong because $P(F \wedge G) = P(F) P(G)$ if they are independent, but in this case they are not. In fact $P(A > B \wedge B > C) = \frac{1}{3}$ because this is saying $B$ is the middle one out of 3 independent identically distributed variables.

Finally, even if all these errors are fixed and you somehow calculate $P(E | F \wedge G)$, this still only gives the probability that $A > 13$ conditioned on $A$ being the max. You have to further argue by symmetry that the cases of $B$ or $C$ being max are equivalent before you can conclude that this is the right answer. (The symmetry argument does apply in this case, so I'm just pointing out you need it.)