Some probability problems

**Dr. Jekyll** · April 19th 2010, 07:36 AM

Hello everybody!

I don't have right answers for the problems below, so I would really appreciate if someone could check my answers. I'm not sure if any of this is right.

1. We have 25 products, 5 of which are bad. What is the probability of
a) choosing one bad product?
b) choosing (any) five products?
c) choosing five products with exactly two bad products?
(Products are chosen randomly.)

a) $\frac{5}{25}=\frac{1}{5}$

b) $\frac{5}{25}=\frac{1}{5}$

c) $\frac{20}{25} \cdot \frac{19}{24} \cdot \frac{18}{23} \cdot \frac{5}{22} \cdot \frac{4}{21}=0.0215$

2. We have 8 products, 3 of which are bad. What is the probability of choosing either BGG or GBG (B - bad product, G - good product)?

$\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} \cdot 2=0.3571$

3. We have 7 good and 4 bad products. What is the probability of
a) choosing three good products?
b) choosing exactly two good products and one bad product?
c) choosing three products with at least one bad product?

a) $\frac{7}{11} \cdot \frac{6}{10} \cdot \frac{5}{9}=0.2121$

b) $\frac{7}{11} \cdot \frac{6}{10} \cdot \frac{4}{9}=0.1697$

c) $\frac{4}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} + \frac{4}{11} \cdot \frac{3}{10} \cdot \frac{7}{9} + \frac{4}{11} \cdot \frac{3}{10} \cdot \frac{2}{9}=0.2788$

**BabyMilo** · April 20th 2010, 04:23 AM

Originally Posted by Dr. Jekyll

Hello everybody!

I don't have right answers for the problems below, so I would really appreciate if someone could check my answers. I'm not sure if any of this is right.

1. We have 25 products, 5 of which are bad. What is the probability of
a) choosing one bad product?
b) choosing (any) five products?
c) choosing five products with exactly two bad products?
(Products are chosen randomly.)

a) $\frac{5}{25}=\frac{1}{5}$

b) $\frac{5}{25}=\frac{1}{5}$

c) $\frac{20}{25} \cdot \frac{19}{24} \cdot \frac{18}{23} \cdot \frac{5}{22} \cdot \frac{4}{21}=0.0215$

a)&b) i believe it's correct.

but c) $\frac{5C2*20C3}{25C5} = 0.215 (3s.f.)$

**BabyMilo** · April 20th 2010, 04:32 AM

Originally Posted by Dr. Jekyll

Hello everybody!

2. We have 8 products, 3 of which are bad. What is the probability of choosing either BGG or GBG (B - bad product, G - good product)?

$\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} \cdot 2=0.3571$

I believe your answer is right.

but it's not times 2

it should be $\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} + \frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6}$

**BabyMilo** · April 20th 2010, 04:42 AM

Originally Posted by Dr. Jekyll

Hello everybody!

3. We have 7 good and 4 bad products. What is the probability of
a) choosing three good products?
b) choosing exactly two good products and one bad product?
c) choosing three products with at least one bad product?

a) $\frac{7}{11} \cdot \frac{6}{10} \cdot \frac{5}{9}=0.2121$

b) $\frac{7}{11} \cdot \frac{6}{10} \cdot \frac{4}{9}=0.1697$

c) $\frac{4}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} + \frac{4}{11} \cdot \frac{3}{10} \cdot \frac{7}{9} + \frac{4}{11} \cdot \frac{3}{10} \cdot \frac{2}{9}=0.2788$

a) is correct.

b) should be $\frac{7C2*4C1}{11C3}$

c) should be 1-(Probability of no bad product)
= $1-\frac{7C3}{11C3}$
=0.788(3s.f.)

**Dr. Jekyll** · April 20th 2010, 05:39 AM

Originally Posted by BabyMilo

I believe your answer is right.

but it's not times 2

it should be $\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} + \frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6}$

Well, isn't it the same thing?

Originally Posted by BabyMilo

a) is correct.

b) should be $\frac{7C2*4C1}{11C3}$

c) should be 1-(Probability of no bad product)
= $1-\frac{7C3}{11C3}$
=0.788(3s.f.)

Of course! Order in which we choose the products is not relevant, that's why we use binomial coefficients (number of combinations), am I right?

There may be some more questions.

Thanks a lot!

**Archie Meade** · April 20th 2010, 06:27 AM

Originally Posted by Dr. Jekyll

Hello everybody!

I don't have right answers for the problems below, so I would really appreciate if someone could check my answers. I'm not sure if any of this is right.

1. We have 25 products, 5 of which are bad. What is the probability of
a) choosing one bad product?
b) choosing (any) five products?
c) choosing five products with exactly two bad products?
(Products are chosen randomly.)

a) $\frac{5}{25}=\frac{1}{5}$

b) $\frac{5}{25}=\frac{1}{5}$ This precedes part c) and is referring to the probability of picking a particular group of 5

$\color{blue}\frac{1}{\binom{25}{5}}$

c) $\frac{20}{25} \cdot \frac{19}{24} \cdot \frac{18}{23} \cdot \frac{5}{22} \cdot \frac{4}{21}=0.0215$ this is the probability that the last 2 picked were bad,

which equals the probability of any 2 of the 5 being bad, those possibilities must be summed.

2. We have 8 products, 3 of which are bad. What is the probability of choosing either BGG or GBG (B - bad product, G - good product)?

$\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} \cdot 2=0.3571$

3. We have 7 good and 4 bad products. What is the probability of
a) choosing three good products?
b) choosing exactly two good products and one bad product?
c) choosing three products with at least one bad product?

a) $\frac{7}{11} \cdot \frac{6}{10} \cdot \frac{5}{9}=0.2121$

b) $\frac{7}{11} \cdot \frac{6}{10} \cdot \frac{4}{9}=0.1697$ this is the probability that only the last one chosen is bad,

we need to sum the probabilities of only the 1st, only the 2nd or only the 3rd is bad.

c) $\frac{4}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} + \frac{4}{11} \cdot \frac{3}{10} \cdot \frac{7}{9} + \frac{4}{11} \cdot \frac{3}{10} \cdot \frac{2}{9}=0.2788$

This is the probability that the 1st is bad or the 1st and 2nd are bad or all 3 are bad.
It doesn't account for only the 2nd being bad, only the last being bad, only the last 2 being bad, the 1st and last being bad.

Hi Dr Jekyll,

some additional feedback on your work and why it's simplest to calculate
as BabyMilo showed.

**downthesun01** · April 21st 2010, 02:39 AM

3. We have 7 good and 4 bad products. What is the probability of
a) choosing three good products?
b) choosing exactly two good products and one bad product?
c) choosing three products with at least one bad product?

I've been looking at this problem 3. Maybe I'm wrong but wouldn't it be easiest to just use a hypergeometric distribution?

For example:
A)
x= the number of good products
X~hypergeometric(N=11, n=3, k=7)
pr(X=3)

$\frac{(^{k}_{x})(^{N-k}_{n-x})}{(^{N}_{n})}$

$\frac{(^{7}_{3})(^{4}_{0})}{(^{11}_{3})}\approx 0.2121$

B)
x= the number of good products
X~hypergeometric(N=11, n=3, k=7)
pr(X=2)

$\frac{(^{k}_{x})(^{N-k}_{n-x})}{(^{N}_{n})}$

$\frac{(^{7}_{2})(^{4}_{1})}{(^{11}_{3})}\approx 0.5091$

C) x= the number of bad products
X~hypergeometric(N=11, n=3, k=4)

pr(X $\geq 1$ )= $\sum ^{3}_{x=1}$ $\frac{(^{k}_{x})(^{N-k}_{n-x})}{(^{N}_{n})}$ = $\frac{(^{4}_{1})(^{7}_{2})}{(^{11}_{3})}$ + $\frac{(^{4}_{2})(^{7}_{1})}{(^{11}_{3})}$ + $\frac{(^{4}_{3})(^{7}_{0})}{(^{11}_{3})} \approx 0.7879$

Maybe I'm wrong, but I believe that is the proper way these problems. Any insight into what I'm doing right or wrong would be awesome.

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