Four probabality Qs: multiple-choice test, buying puppy, etc

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deepika12

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Jun 1, 2007
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5
I JUSt cannot solve these problems. HELP!!!

1) A pet store has 7 puppies, including 4 poodles, 2 terriers, and one retriever. If Rebecca and Aaron (in that order) each select one puppy at random, with replacement (they may both select the same one) find the probability of them both selecting a retriever?

2) Yesa Brill is taking a 10 question multiple choice test for which each question has three answer choices, only one of which is correct. Yesha decides on answers by rolling a fair die and marking the first answer choice if the die shows 1 or 2, the second if it shows 3 or 4, and the 3rd if it shows 5 or 6. Find the probability of exactly 4 correct answers?

3) In a certain state, it has been shown that only 50% of the high school graduates who are capable of college work actually enroll in college. Find the probability that amoung the 9 capable high school students in this state, each of the following numbers will enroll in college? EXACTLY 4.

4) If 30% of all workers in a certain industry are union members, and workers in this industry are selected successively at random, find the probability that the first union member to occur will be on the third selection?
 
You may not have been able to finish ("solve") these exercises, but I'm sure that you've been able at least to get started, and that you have some thoughts, right? :wink:

Please reply with all of that information, so the tutors can follow your lead (rather than accidentally heading off into techniques you maybe haven't seen) and thus help you get un-stuck.

Thank you! :D

Eliz.
 
According to me the answer for
1. 1/196

3.0.24609375


Are these correct.
I am still trying to figure out the rest.
 
The second one is a binomial also.

\(\displaystyle \L\\C(10,4)(\frac{1}{3})^{4}(\frac{2}{3})^{6}\)

What is 3.0.24609... supposed to mean, may I ask?.
 
I mean Answer for 3rd. is 0.24609375
Is that correct ?
and is the answer for 1st : 1/196 ?
 
I get .227 for the third one.

The first one looks like it would be (1/7)(1/7)=1/49.

There's only one retriever in 7 dogs.
 
How do u get .227 for 3rd
I got by
C(9,4)(0.05^4)(0.05^5)


And oh yeah... I was making the mistake of considering 7 pupies + 4 labs + and so on.... I missed that in all there r just 7 pups which include....

Thanks.
A lot.
 
For number 3, where are you getting 9?. There are 10 questions.

Also, where does the .05 comes from?
 
3) In a certain state, it has been shown that only 50% of the high school graduates who are capable of college work actually enroll in college. Find the probability that amoung the 9 capable high school students in this state, each of the following numbers will enroll in college? EXACTLY 4.

I get 0.05 from 50%
and its 9 students sample.
 
Re: Four probabality Qs: multiple-choice test, buying puppy,

Hello, deepika12!

1) A pet store has 7 puppies, including 4 poodles, 2 terriers, and one retriever.
If Rebecca and Aaron (in that order) each select one puppy at random, with replacement
(they may both select the same one), find the probability of them both selecting a retriever?
Click to expand...

\(\displaystyle P(\text{Rebecca selects the retriever}) \:=\:\frac{1}{7}\)
\(\displaystyle P(\text{Aaron selects the retriever}) \:=\:\frac{1}{7}\)
\(\displaystyle P(\text{both select the retriever}) \:=\:\left(\frac{1}{7}\right)\cdot\left(\frac{1}{7}\right) \:=\:\L\fbox{\frac{1}{49}}\)


2) Yesha Brill is taking a 10 question multiple-choice test
for which each question has three answer choices, only one of which is correct.
Yesha decides on answers by rolling a fair die
and marking the first answer choice if the die shows 1 or 2,
the second if it shows 3 or 4, and the 3rd if it shows 5 or 6.
Find the probability of exactly 4 correct answers?
Click to expand...

We have: \(\displaystyle \:p(\text{right}) \:=\:\frac{1}{3},\;P(\text{wrong}) \:=\:\frac{2}{3}\)

THen: \(\displaystyle \:p(\text{4 right, 5 wrong}) \:=\:\L{10\choose4}\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^5 \;=\;210\cdot\frac{64}{59049} \:=\:\fbox{0.22760758}\)


3) In a certain state, it has been shown that only 50% of the high school graduates
who are capable of college work actually enroll in college.
Find the probability that amoung the 9 capable high school students in this state,
exactly four will enroll in college?
Click to expand...

We have: \(\displaystyle \:p(\text{college})\:=\:0.5,\;P(\text{not}) \:=\:0.5\)

Then: \(\displaystyle \:p(\text{4 college, 5 not}) \:=\:\L{9\choose4}(0.5)^4(0.5)^5\;=\;\L\fbox{0.24609375}\)


4) If 30% of all workers in a certain industry are union members,
and workers in this industry are selected successively at random,
find the probability that the first union member to occur will be on the third selection?
Click to expand...

We have: \(\displaystyle \:p(U) \:=\:0.3,\;P(N) \:=\:0.7\)

The first two must be non-union workers: \(\displaystyle \:p(NN) \:=\:(0.7)^2\)
The third must be a union worker: \(\displaystyle \:p(U) \:=\:0.3\)

Therefore: \(\displaystyle \:p(NNU) \:=\:(0.7)^2(0.3) \:=\:\L\fbox{0.147}\)

 
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