# Math Help - On which probability.

1. ## On which probability.

Hello.

A teacher gives to a class of 10 matters and says that they will become to the examination
the 5 ones ( 5 out of 10 ) that have been occasionally chosen.
The student x has learned 7 ( 7 out of 10 ) of them. On which probability in the examination
all 5 matters are those the student x has learned?

Keyword is multiplication, I believe.

There should be 2 ways to answer the first question; right or wrong.
2*

There should be 2 ways to answer the second question; right or wrong.
2*2

Third question
2*2 *2

Fourth question
2*2 *2 *2

Fifth question
2*2 *2 *2 * 2

Sixth question
2*2 *2 *2 * 2 *2

Seventh question
2*2 *2 *2 * 2 *2 * 2

Eight question
2*2 *2 *2 * 2 *2 * 2 * 2

Ninth question
2*2 *2 *2 * 2 *2 * 2* 2 * 2

Tenth question
2*2 *2 *2 * 2 *2 * 2 * 2 * 2 * 2

I believe that probability ( it is not2^10 )

is 2^5 = 1/32. ?

2. ## Re: On which probability.

Originally Posted by stevetall
Hello.

A teacher gives to a class of 10 matters and says that they will become to the examination
the 5 ones ( 5 out of 10 ) that have been occasionally chosen.
The student x has learned 7 ( 7 out of 10 ) of them. On which probability in the examination
all 5 matters are those the student x has learned?

Keyword is multiplication, I believe.

There should be 2 ways to answer the first question; right or wrong.
2*

There should be 2 ways to answer the second question; right or wrong.
2*2

Third question
2*2 *2

Fourth question
2*2 *2 *2

Fifth question
2*2 *2 *2 * 2

Sixth question
2*2 *2 *2 * 2 *2

Seventh question
2*2 *2 *2 * 2 *2 * 2

Eight question
2*2 *2 *2 * 2 *2 * 2 * 2

Ninth question
2*2 *2 *2 * 2 *2 * 2* 2 * 2

Tenth question
2*2 *2 *2 * 2 *2 * 2 * 2 * 2 * 2

I believe that probability ( it is not2^10 )

is 2^5 = 1/32. ?
What is the probability that the first question is one of the seven? Given the first was one of the seven, what is the probability that the second is one of the seven. ....

CB

3. ## Re: On which probability.

Let's try again.

5 * 0,7 + 5 * 0,7 + 5 * 0,7 + 5 * 0,7 + 5 * 0,7 5 * 0,7 + 5 * 0,7 / 5 = 0,49

4. ## Re: On which probability.

Originally Posted by stevetall
Let's try again.

5 * 0,7 + 5 * 0,7 + 5 * 0,7 + 5 * 0,7 + 5 * 0,7 5 * 0,7 + 5 * 0,7 / 5 = 0,49
What do you think the above calculation represents and why would that be an answer to the question.

Have you considered answering the questions I asked in my previous post?

CB

5. ## Re: On which probability.

Sorry. Yes, of course I tried to follow Your advice. But I think more and try again. It is not easy.

6. ## Re: On which probability.

first I thought that just

1/7 * 7, because it is always 1/7 ( on questions ) and there are 7 questions that student x has learned.

But it can't be like that.

So this is the best I can do ( I know
that this may be incorrect and something is possibly missing)
Maybe more multiplication is needed?

first question is 1/7
second is 1/7 x 1 /7 = 1 out of 7^2
third is 3 * 1/7 = 1 out of 7^3
fourth is 4 * 1/7 = 1 out of 7^4
fifth is 5 * 1/7 = 1 out of 7^5
sixth is 6 * 1/7 = 1 out of 7^6
seventh is 7 * 1/7 = 1 out of 7^7

7. ## Re: On which probability.

Is my solution wrong / & something is missing? - I don't know. I tried to follow advice.

8. ## Re: On which probability.

Originally Posted by stevetall
Is my solution wrong / & something is missing? - I don't know. I tried to follow advice.
The probability that the first question is one of your 7 is 7/10 (favourable cases over all possible since these are all equally likely)

Given the first question is one of the 7 the probability that the second is one of the 7 is 6/9 (one of the seven gone leaving 6 favourable cases and a total of 9)

etc.

CB

9. ## Re: On which probability.

Hi. I try again.

1. 7/10 = 0,7
2. 6/9 = 0,67
3. 5/8 = 0,625
4. 4/7 = 0,57
5. 3/6 = 0,5
6. 2/5 = 0,4
7. 1/4 = 0,25

And now, I believe, we must do this;

0,7 + 0,67 + 0,625 + 0,57 + 0,5 + 0,4 + 0,25 / 7 = 9,95 %
but can this be a right probability?

10. ## Re: On which probability.

Originally Posted by stevetall
Hi. I try again.

1. 7/10 = 0,7
2. 6/9 = 0,67
3. 5/8 = 0,625
4. 4/7 = 0,57
5. 3/6 = 0,5
6. 2/5 = 0,4
7. 1/4 = 0,25

And now, I believe, we must do this;

0,7 + 0,67 + 0,625 + 0,57 + 0,5 + 0,4 + 0,25 / 7 = 9,95 %
but can this be a right probability?
1. You are only choosing 5 questions

2. You multiply the probabilities together because you want all the events to occur

CB

11. ## Re: On which probability.

Hello, stevetall!

I hope I interpreted the question correctly.

A teacher gives his class 10 questions says that 5 of them will be on the exam.
A student has mastered 7 of the 10 questions.
What is the probability that the student knows all 5 questions on the exam?

. . $\begin{array}{ccc}\text{Question} & \text{Prob.} \\ \hline \\[-4mm] 1 & \frac{7}{10} \\ \\[-4mm] 2 & \frac{6}{9} \\ \\[-4mm] 3 & \frac{5}{8} \\ \\[-4mm] 4 & \frac{4}{7} \\ \\[-4mm] 5 & \frac{3}{6} \end{array}$

$\text{Probability} \;=\;\frac{7}{10}\cdot\frac{6}{9}\cdot\frac{5}{8} \cdot \frac{4}{7}\cdot\frac{3}{6} \;=\;\frac{1}{12}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

An alternate approach . . .

There are: . $_{10}C_5 \:=\:\frac{10!}{5!\,5!} \:=\: 252$ possible exams.
The student will know all 5 questions in: . $_7C_5 \:=\:\frac{7!}{5!\,2!} \:=\:21$ of them.

Therefore: . $\text{Probability} \;=\;\frac{21}{252} \;=\;\frac{1}{12}$

12. ## Re: On which probability.

Soroban, yes, I meant just like that. Thank you!