Bayes Theorem

**axa121** · October 18th 2009, 06:57 AM

Hi there could anyone help me out with the following question please:

A plane crash could be caused by 1 or 4 of the following events.

H1 = faulty motor P(H1) = 0.2
H2 = bomb in the plane P(H2) = 0.4
H3 = missile that hit the plane P(H3) = 0.3
H4 = fault in the computer system P(H4) = 0.1

The investigation has shown that the crash was accompanied by the event A = fuel ignition. It is known that:
P(A l H1 ) = 0.7;
P(A l H2 ) = 0.9
P(A l H3 ) = 0.8
P(A l H4 ) = 0

Calculate P(H2 l A )

So i tried using the Bayes theorem but i got a probability of over 1. Could anyone help me out please.

**CaptainBlack** · October 18th 2009, 07:45 AM

Originally Posted by axa121

Hi there could anyone help me out with the following question please:

A plane crash could be caused by 1 or 4 of the following events.

H1 = faulty motor P(H1) = 0.2
H2 = bomb in the plane P(H2) = 0.4
H3 = missile that hit the plane P(H3) = 0.3
H4 = fault in the computer system P(H4) = 0.1

The investigation has shown that the crash was accompanied by the event A = fuel ignition. It is known that:
P(A l H1 ) = 0.7;
P(A l H2 ) = 0.9
P(A l H3 ) = 0.8
P(A l H4 ) = 0

Calculate P(H2 l A )

So i tried using the Bayes theorem but i got a probability of over 1. Could anyone help me out please.

Bayes' theorem for this says:

$p(H_2|A)=\frac{p(A|H_2)p(H_2)}{p(A)}$

$p(A)=p(A|H_1)p(H_1)+p(A|H_2)p(H_2)+p(A|H_3)p(H_3)+ p(A|H_4)p(H_4)$

and if you do the arithmetic right $p(H_2|A)<0.5$ .

CB

**Soroban** · October 18th 2009, 08:28 AM

Hello, axa121!

You left out a piece of data . . . I'll make one up.

A plane crash could be caused by 1 or 4 of the following events:

. . $\begin{array}{cccccccc} \text{faulty motor} & P(\text{motor}) &=& 0.2 \\ \text{bomb} & P(\text{bomb}) &=& 0.4 \\ \text{missible} & P(\text{missile}) &=& 0.3 \\ \text{computer} & P(\text{computer} &=& 0.1 \end{array}$

The crash was accompanied by the event $\text{fuel ignition}$ .

It is known that:

. . $\begin{array}{ccccc} P(\text{fuel }|\text{ motor}) &=& 0.7 & (1)\\ P(\text{fuel }|\text{ bomb}) &=& 0.9 & (2)\\ P(\text{fuel }|\text{ missile}) &=& 0.8 & (3)\\ P(\text{fule }|\text{ computer}) &=& 0.6 & (4)\end{array}$

Calculate $P(\text{bomb }|\text{ fuel})$

We want: . $P(\text{bomb }|\text{ fuel}) \;=\; \frac{P(\text{bomb} \wedge \text{fuel})} {P(\text{fuel})}$ .(a)

From (1), we have: . $P(\text{fuel }|\text{ motor}) \:=\:\frac{P(\text{fuel}\wedge\text{motor})}{P(\te xt{motor})} \:=\:0.7 \quad\Rightarrow\quad \frac{P(\text{fuel}\wedge\text{motor})}{.2} \:=\:0.7$
. . Hence: . $P(\text{fuel}\wedge\text{motor}) \:=\:0.14$

From (2), we have: . $P(\text{fuel }|\text{ bomb}) \;=\;\frac{P(\text{fuel}\wedge\text{bomb})}{P(\tex t{bomb})} \:=\:0.9\quad\Rightarrow\quad \frac{P(\text{fuel}\wedge\text{bomb})}{0.4} \:=\:0.9$
. . Hence: . $P(\text{fuel}\wedge\text{bomb}) \:=\:0.36$ .(b)

From (3), we have: . $P(\text{fuel }|\text{ missle}) \;=\;\frac{P(\text{fuel}\wedge\text{missile})}{P(\ text{missible})} \:=\:0.8 \quad\Rightarrow\quad \frac{P(\text{fuel}\wedge\text{missile})}{0.3} \:=\:0.8$
. . Hence: . $P(\text{fuel}\wedge\text{missile}) \;=\;0.24$

Fromn (4), we have: . $P(\text{fuel }|\text{ computer}) \:=\:\frac{P(\text{fuel}\wedge\text{computer})}{P( \text{computer})} \:=\:0.6 \quad\Rightarrow\quad \frac{P(\text{fuel}\wedge\text{computer})}{0.1} \:=\:0.6$
. . Hence: . $P(\text{fuel}\wedge\text{computer}) \:=\:0.06$

Then: . $P(\text{fuel}) \:=\:0.14 + 0.36 + 0.24 + 0.06 \;=\;0.80$ .p(c)

Substitute (b) and (c) into (a): . $P(\text{bomb }|\text{fuel}) \;=\;\frac{P(\text{bomb}\wedge\text{fuel})}{P(\tex t{fuel})} \;=\;\frac{0.36}{0.80} \;=\;\boxed{0.45}$

**axa121** · October 18th 2009, 08:29 AM

Thanks, the way I had done it I got something like 67/35.

The thing I am finding more difficlult with the Bayes Theorem is splitting the outcomes into partitions, but this question had already done so. I''l practice a few more examples.

**axa121** · October 18th 2009, 08:41 AM

Soroban in the question it says P ( A l H4 ) = 0

Thanks for the help though.

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