# Probability Problem

• Jun 7th 2009, 08:05 AM
AlgebraicallyChallenged
Probability Problem
Dear Forum , I am having trouble figuring out the following , any help would be appreciated...
A shipment of 60 inexpensive watches, including 9 that are defective, is sent to a department store. The receiving department selects 10 at random (from the 60 received) for testing, and rejects the entire shipment if 1 or more in the sample is found defective. What is the probability that the shipment will be rejected?

thanks - AC- (Wondering)
• Jun 7th 2009, 08:45 AM
apcalculus
Quote:

Originally Posted by AlgebraicallyChallenged
Dear Forum , I am having trouble figuring out the following , any help would be appreciated...
A shipment of 60 inexpensive watches, including 9 that are defective, is sent to a department store. The receiving department selects 10 at random (from the 60 received) for testing, and rejects the entire shipment if 1 or more in the sample is found defective. What is the probability that the shipment will be rejected?

thanks - AC- (Wondering)

Let X be the number of defective units in the sample of 10.

P(X= or > 1)= P(X=1) + P(X=2) +...+ P(X=9) since 9 is the maximum number of defective units in the shipment.

P(X=1) is the probability of getting EXACTLY one defective unit, then all the other nine work.

= (9/60) (51/59)(50/58) (49/57) continue until you have 10 terms.
Doing this for all nine cases is a lot of work so we write:

P(X>=1) = 1 - P(X <1)

P(X < 1) = P(X=0) which is the probability of getting 0 defective units. This means all 10 units chosen at random will work:

51/60 is the probability of the first one working
50/59 is the probability of the second one working

... write them all out, and multiply them since we are assuming independence.

Good luck!
• Jun 7th 2009, 10:21 AM
Amer
Quote:

Originally Posted by AlgebraicallyChallenged
Dear Forum , I am having trouble figuring out the following , any help would be appreciated...
A shipment of 60 inexpensive watches, including 9 that are defective, is sent to a department store. The receiving department selects 10 at random (from the 60 received) for testing, and rejects the entire shipment if 1 or more in the sample is found defective. What is the probability that the shipment will be rejected?

thanks - AC- (Wondering)

I have a long solution if the receiving find the first defective they will reject so the probability is
(d)+(n,d)+(n,n,d)+(n,n,n,d)+(n,n,n,n,d)+(n,n,n,n,n ,d)+........(n,n,n,n,n,n,n,n,n,d) that is it

so the probability that they will find the defective from the first one selected then they will reject so the probability of this is

$\frac{9}{60}$

or they will find the defective one from the second selection so the probability will be

$\left(\frac{51}{60}\right)\left(\frac{9}{59}\right )$

or they will find the defective one from the third selection so the probability will be

$\left(\frac{51}{60}\right)\left(\frac{50}{59}\righ t)\left(\frac{9}{58}\right)$

or from the fourth selection

$\left(\frac{51}{60}\right)\left(\frac{50}{59}\righ t)\left(\frac{49}{58}\right)\left(\frac{9}{57}\rig ht)$

or from the fifth selection

$\left(\frac{51}{60}\right)\left(\frac{50}{59}\righ t)\left(\frac{49}{58}\right)\left(\frac{48}{57}\ri ght)\left(\frac{9}{56}\right)$
.
.
.
.
.
find them all then find the sum of them all or equal add
• Jun 7th 2009, 10:25 AM
Amer
Quote:

Originally Posted by apcalculus
Let X be the number of defective units in the sample of 10.

P(X= or > 1)= P(X=1) + P(X=2) +...+ P(X=9) since 9 is the maximum number of defective units in the shipment.

P(X=1) is the probability of getting EXACTLY one defective unit, then all the other nine work.

= (9/60) (51/59)(50/58) (49/57) continue until you have 10 terms.
Doing this for all nine cases is a lot of work so we write:

P(X>=1) = 1 - P(X <1)

P(X < 1) = P(X=0) which is the probability of getting 0 defective units. This means all 10 units chosen at random will work:

51/60 is the probability of the first one working
50/59 is the probability of the second one working

... write them all out, and multiply them since we are assuming independence.

$1-\frac{\binom{51}{10}}{\binom{60}{10}}$