# Thread: Simple probability question...

1. ## 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.

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

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

= ......

3. 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 ...

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

4. 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 ...

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:

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

$\displaystyle = 2 \left( \frac{3}{10}\right) \left( \frac{7}{9}\right) + \left( \frac{3}{10}\right) \left( \frac{2}{9}\right)$
(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)

= ......

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

Thanks again.

6. 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).

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