Simple probability question...

• Apr 21st 2008, 04:16 AM
struck
Simple probability question...
Out of 20 components, 3 are defective. Two components are chosen at random from these 20 components. Determine the probability that they will both be defective.

I thought these are two independent events occurring in any order, thus:

P(E1 and E2 in any order) = P(E1 then E2) + P(E2 then E1)

But that doesn't give me the right answer. :(
• Apr 21st 2008, 04:32 AM
mr fantastic
Quote:

Originally Posted by struck
Out of 20 components, 3 are defective. Two components are chosen at random from these 20 components. Determine the probability that they will both be defective.

I thought these are two independent events occurring in any order, thus:

P(E1 and E2 in any order) = P(E1 then E2) + P(E2 then E1)

But that doesn't give me the right answer. :(

$\frac{ {3 \choose 2} {17 \choose 0} }{{20 \choose 2}}$

$= \left( \frac{3}{20} \right) \left( \frac{2}{19} \right)$
(unsurprisingly)

= ......
• Apr 21st 2008, 05:50 AM
struck
Thanks.. Another related question.

The letters of word STATISTICS (10) are written, one letter on each of ten cards. The cards are placed face downwards on a table and shuffled and two cards are turned over together. What is the probability that:
a) one, at least, is a vowel? (A or I)

I think here:

P(E1) = 3/10
P(E2) = 2/9

But I don't get how to solve it ... (Doh)

Ohh, and in simple terms please. It's for GCSE and I have to learn this part (probability) on my own.
• Apr 21st 2008, 06:13 AM
mr fantastic
Quote:

Originally Posted by struck
Thanks.. Another related question.

The letters of word STATISTICS (10) are written, one letter on each of ten cards. The cards are placed face downwards on a table and shuffled and two cards are turned over together. What is the probability that:
a) one, at least, is a vowel? (A or I)

I think here:

P(E1) = 3/10
P(E2) = 2/9

But I don't get how to solve it ... (Doh)

Ohh, and in simple terms please. It's for GCSE and I have to learn this part (probability) on my own.

One vowel or two vowels:

$\frac{{3 \choose 1} {7 \choose 1}}{ {10 \choose 2} } + \frac{{3 \choose 2} {7 \choose 0}}{ {10 \choose 2} }$

$= 2 \left( \frac{3}{10}\right) \left( \frac{7}{9}\right) + \left( \frac{3}{10}\right) \left( \frac{2}{9}\right)$
(unsurprisingly - note that the 2 in the first term occurs because you can get the one vowel on either the first or the second turn over)

= ......
• Apr 22nd 2008, 02:31 AM
struck
Although I got an answer, I still don't seem to get the 'why' - specifically the multiplier 2 in there.

Thanks again.
• Apr 22nd 2008, 03:00 AM
mr fantastic
Quote:

Originally Posted by struck
Although I got an answer, I still don't seem to get the 'why' - specifically the multiplier 2 in there.

Thanks again.

1 vowel in two turns: V V' or V' V.

Pr(V V') = (3/10) (7/9).
Pr(V' V) = (7/10) (3/9) = Pr(V V').

So Pr(V V' or V' V) = Pr(V V') + Pr(V' V) = 2(3/10)(7/9).
• Apr 22nd 2008, 03:04 AM
mr fantastic
Quote:

Originally Posted by struck
Although I got an answer, I still don't seem to get the 'why' - specifically the multiplier 2 in there.

Thanks again.

Alternatively, $\frac{{3 \choose 1} {7 \choose 1}}{ {10 \choose 2} }$ simplifies to $2 \left( \frac{3}{10}\right) \left( \frac{7}{9}\right)$ and the previous post shows why this is unsurprising.