Lino's Essay Errors

If there are nine pages in Lino's essay and seven correct pages in Lino's essay, then the probability that the teacher does NOT pick one of the pages with errors is $\frac {7}{9}$ . Now, there are eight pages and six of them have errors. The probability that the teacher does not pick a page with errors here is $\frac {6}{8}$ . We repeat this for the last time the teacher picks a page, and we get $\frac {5}{7}$ .

Since these events are independent, the probability that the teacher doesn't read any of the pages with errors can be expressed as $\frac {7}{9} \cdot \frac {6}{8} \cdot \frac {5}{7}$ which is $\frac {5}{12}$ . $5 + 12 = 17.$

The only way in which the teacher can read the pages without any errors is that she reads from the 7 pages that have no error in them. Since she randomly chose the pages, the number of choices possible is $7 \choose 3$ . The total number of combinations possible, including the pages with errors, is $9 \choose 3$ . Therefore, the answer is $\frac {7 \choose 3}{9 \choose 3}$ , which equals 5+12=17.

In the essay, there would be 7 pages with no errors

The number of combinations of which all 3 pages are correct is: $7 \times 6 \times5$ = 210

The number of combinations which could have been chosen is: $9 \times 8 \times7$ = 504

Thus, the probability would be: $\frac{210}{504}$

Simplify : $\frac{5}{12}$

$5+12=17$

There are nCr(9, 3)=84 ways the teacher can select 3 pages randomly. There are nCr(7, 3)=35 ways to select 3 pages randomly from the set of pages with no errors. Thus, the desired probability is 35/84=5/12. Thus, the answer is 5+12=17

3 pages are gonna be read and none of them should contain error

the probability that the first page chosen is without error is \frac{7}{9}\ The probability that the second page chosen is without error is \frac{7-1}{9-1}\, because the first page is not to be chosen again The probability that the third page chosen is without error is \frac{7-2}{9-2}\ because the first and second page is not to be chosen again multiplying the 3 fractions will give you \frac{5}{12}\, 5+12 = 17

For the teacher to not choose either page 2 or page 7, he or she must only read pages other than page 2 or 7. There are 7 pages in which there are no errors, thus the number of ways the teacher can read any of these 3 pages is {7 \choose 3} = 35. The total number of ways the teacher can read any of the 3 pages is {9 \choose 3}= 84. Thus, the probability that the teacher did not read any of the pages is 35/84 = 5/12. Thus a=5, b=12 and a+b = 17.

First method

The teacher can choose 3 pages in $\binom{9}{3}$ ways, but we want him to choose only the pages whitout errors, so he must choose $\binom{7}{3}$ .

The quotient is $\frac{ \binom{7}{3} }{ \binom{9}{3} } = \frac{5}{12}$ , hence the solution is $5 + 12 = 17$

Second method

We can construct a binary tree, so for the first children we have a probability of $\frac{7}{9}$ , in the second one we have a probability of $\frac{6}{8}$ and in the last one $\frac{5}{7}$ . After multiplying all numbers we have $\frac{7}{9} \cdot \frac{6}{8} \cdot \frac{5}{7} = \frac{5}{12}$ , hence the solution is $5 + 12 = 17$

there are 84 ways in which the teacher read 3 of the 9 pages, that would total the forms and if you do not want to read pages 2 and 7, would have to choose 3 the 7 pages remaining: 35 forms would exist entoces what I ask for is 35/84 = 5/12, with 5 and 12 relatively prime 5 +12 = 17

The probability that the teacher will read the first page with no errors is 7/9. Next page is 6/8 (because she has already used up one page), and the next probability is 5/7. Multiplying together, we get $\frac{7\cdot 6 \cdot 5}{9 \cdot 8 \cdot 7}$ . After cancelling common factors, we are left with $\frac{5}{3 \cdot 4}$ , so a+b=5+12=17.

total number of possible ways to choose from 9 pages only 3 pages is = (9 * 8 * 7)/ 3!

and the number of possible ways not to choose pages 2 and 7 are same as choosing 3 pages from 7 pages = (7 * 6 * 5/ 3!)

SO PROBABILITY IS (7 * 6 * 5/ 3!)/(9 * 8 * 7)/ 3! = 5/12 = 17

First, consider the total number of ways the teacher can choose three pages out of nine choices (do note that 0rder is not important because she chooses randomly and order isn't considered). That is,

$_9C_3 = 84$

Second, consider the restriction given. It is indicated that two pages have errors and that the teacher will not be able to choose them. So that means, the teacher can choose the three pages out of seven choices. That is,

$_7C_3 = 35$

The probability that the teacher doesn't read the pages with error is equal to

$\dfrac{_7C_3}{_9C_3}=\dfrac{35}{84} =\dfrac{5}{12}$

Since $a = 5$ and $b = 12$ , then $a + b =17$ .

The total number of ways the teacher can choose the 3 pages to read is $\binom{9}{3}$ . There are $9-2=7$ pages without errors, so the total number of ways the teacher can choose 3 pages with no error is $\binom{7}{3}$ , so the probability that she did not find an error is $\frac{\binom{7}{3}}{\binom{9}{3}} = \frac{5}{12}$ . Therefore, $a + b = 5+12 = 17$ .