1. ## probability

A machine manufactures components. The probability of a faulty
component is 0.04. If we sample 20 components what is the probability
of the machine producing 18 good ones?

need help with this one....its for 7 marks so its givin me a hint it could more diffficult than i expected?

2. Try this, X~Bi(20,0.96) , find P(X=18)

3. Originally Posted by decoy808
A machine manufactures components. The probability of a faulty
component is 0.04. If we sample 20 components what is the probability
of the machine producing 18 good ones?

need help with this one....its for 7 marks so its givin me a hint it could more diffficult than i expected?
Hi decoy808,

the binomial theorem handles these questions quickly.

If you have not covered this,
then here's a way to think it through.

Exactly 18 good ones means 18 good ones and 2 bad ones.

When the machine produced the 20 components,
it could have produced 2 bad ones followed by 18 good ones,
or the first was bad, 2nd one good, 3rd bad, rest good etc

This would produce a long list of possible sequences...

BBGGGGGGGGGGGGGGGGGG
BGBGGGGGGGGGGGGGGGGG
BGGBGGGGGGGGGGGGGGGG

all the way up to

GGGGGGGGGGGGGGGGGGBB

Hence we must either count the number of ways to select 2 positions in 20
or 18 positions in 20 for the good ones,
either way gives the same answer.

We must sum all the seperate probabilities for each sequence,
as these are all the ways the machine could have produced 2 bad ones.

The number of ways is $\binom{20}{2}=\binom{20}{18}$

The probability that 2 bad ones and 18 good ones are together in a sequence is

$0.04^2\ 0.96^{18}$

Therefore the probability of 18 good components out of 20 is $\binom{20}{2}0.04^2\ 0.96^{18}=\binom{20}{18}0.04^2\ 0.96^{18}$

4. thank you for your time and help