# Binomial distribution.

• Jul 6th 2011, 12:33 PM
JWEI
Binomial distribution.
65% of the dishwashers manufactured by a factory do not need repairs for the first three years.If 8 of the dishwashers are to be selected by chance, find the probability
i) all 8 will not need repairs for the first three years
ii) at least two will need repairs within three years

someone can help me solve this question? (Speechless)
• Jul 6th 2011, 12:40 PM
arcketer
Re: Probability Question
For part i.)

You know that 65% of dishwashers do NOT need repairs for 3 years. You select 8 dishwashers at random. Each has 65% chance of not needing repairs. We can extrapolate then. The chance of 1 dishwasher not needing repairs = (0.65). The chance of 2 not needing repairs is (0.65)(0.65). You can continue on from there. Does that make sense?
• Jul 6th 2011, 02:43 PM
mr fantastic
Re: Binomial distribution.
Quote:

Originally Posted by JWEI
65% of the dishwashers manufactured by a factory do not need repairs for the first three years.If 8 of the dishwashers are to be selected by chance, find the probability
i) all 8 will not need repairs for the first three years
ii) at least two will need repairs within three years

someone can help me solve this question? (Speechless)

I re-titled your post "Binomial distribution". I assume the question has come from that part of your work? So, define the random variable. Now, what is p, what is n? What calculation is required? Note that Pr(at least 2) = 1 - Pr(less than 2) ....
• Jul 6th 2011, 02:50 PM
pickslides
Re: Binomial distribution.
Make p is the probability of having a dishwasher that does not need repair. Find where k=8 for

$\displaystyle P(k) = ^nC_k \times p^k(1-p)^{n-k}$
• Jul 6th 2011, 03:44 PM
Plato
Re: Binomial distribution.
Quote:

Originally Posted by pickslides
Make p is the probability of having a dishwasher that does not need repair. Find where k=8 for
$\displaystyle P(k) = ^nC_k \times p^k(1-p)^{n-k}$

May I suggest that the following LaTeX notation is easier to follow.
$$P(k)=\binom{n}{k}p^k(1-p)^{n-k}$$ gives
$P(k)=\binom{n}{k}p^k(1-p)^{n-k}$