# Thread: Simple Probability Problem

1. ## Simple Probability Problem

In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.

i) What % of the town's population is taller than 180 cm?

ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?

Ummm I'm not that good at probability so could someone please give me some help?

For:

i) I did:

Percentage = 100(0.04*0.4 + 0.01*0.6)
= 0.022*100
= 2.2%

ii) I did:

P = 0.022*0.2
= 11/2500

The answers don't look rite though...

2. You have the first part right, for the second one you need to use Baye's theorem.

$
P(woman|taller than 180)= \frac{P(taller than 180|woman)P(woman)}{P(taller than 180|woman)P(woman)+P(taller than 180|man)P(man)}
$

3. Originally Posted by xwrathbringerx
In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.

i) What % of the town's population is taller than 180 cm?

ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?

Ummm I'm not that good at probability so could someone please give me some help?

For:

i) I did:

Percentage = 100(0.04*0.4 + 0.01*0.6)
= 0.022*100
= 2.2% Mr F says: Correct.

ii) I did:

P = 0.022*0.2 Mr F says: Wrong. Where has the 0.2 come from?
= 11/2500

The answers don't look rite though...
ii requires conditional probability.

Here is the magic recipe, the big secret, the great trick, the golden elixir. Read it carefully ........

Set up careful notation and definitions. Carefully define all known probabilities in terms of this notation.

Let T be the event height greater than 180 cm.

You have:

Pr(M) = 2/5, Pr(W) = 3/5,

Pr(T | M) = 1/25, Pr(T | W) = 1/100,

Pr(M and T) = Pr(T| M) Pr(M) = (1/25) (2/5) = 2/125,

Pr(W and T) = Pr(T| W) Pr(W) = (1/100) (3/5) = 3/500,

Pr(T) = Pr(M and T) + Pr(W and T) = 2/125 + 3/500 = 11/500 (= 0.022 as you found).

You require Pr(W | T).

$\Pr(W | T) = \frac{\Pr(W \cap T)}{\Pr(T)} = \frac{3/500}{11/500} = 3/11$.

4. You have the first part right, for the second one you need to use Baye's theorem.
To keep not that messy, Let
W=woman
M=Man
T=taller than 180

$
P(W|T)= \frac{P(T|W)P(W)}{P(T|W)P(W)+P(T|M)P(M)}=\frac{(0. 01)(0.60)}{(0.01)(0.60)+(0.04)(0.40)}
$

$
=\frac{0.006}{0.006+0.016}=\frac{0.006}{0.022}=\fr ac{3}{11}
$