During the halftime break of a football match, the coach plans to make

\(C^3_9=\frac{9!}{3!6!}\);

\(C^3_{11}=\frac{11!}{3!8!}\);

\(\frac{C^3_9}{C^3_{11}}=\frac{9!}{3!6!}*\frac{3!8!}{11!}=\frac{7*8}{10*11}=\frac{28}{55}\).

Hope it helps.

Hi Bunuel here is my approach using probability

1/11 +1/11+1/11 = 3/11 ( its a case when we substitute two forwards and and the third one is anyone from the team)

1-3/11 = 8/11 is number of desired out comes

Now # of desired outcomes divided by number of total outcomes

\(\frac{8}{11} / \frac{55}{1}= \frac{8}{11} *\frac{1}{55}\)

what did i do wrong?

pushpitkc hello thanks for your comments.

can you please tell me wat am I doing wrong when solving this question using probability ?

can you solve it using probability ?

If solving by probability - the easiest way to look at this question is to solve by not including the forwards. Think of solving 3 slots.
First slot = 9/11 (9 players out of 11 can be selected)
Second slot = 8/10 ( minus 1 player since you have selected 1 in the first slot from available / minus 1 player from total available since you selected 1 from the total available)
3rd slot = 7/9 (minus another player since you have selected 2 from the previous two slots of non-forwards / minus another player from 2 the 2 that have been selected from the total players)

the 3 slots will be (9/11) * (8/10) * (7/9) = 28/55

You can also see a pattern where the numerator is reduced by one for each slot with the favorable outcome. The same pattern is seen in the denominator out of the total players on the roster in each slot as well.

11 players ---- 2 forwards ----- 9 non forwards

We want to substitute non forwards :

\(\frac{9}{11} * \frac{8}{10} * \frac{7}{9}\)

\(\frac{28}{55}\)

(C)

Hey hdavies thanks for your comments

but the question is what is the probability that none of the forwards will be substituted?


why do you start with 9/11 and why do you multiply events ? if there is substituon dont we need to add event ?

what would be other way to solve it via probability ?

Out of 11 players, 2 of them are forwards. The question asks for the probability of none of them being chosen. If you remove the 2 forwards from the group, then you will have all the remaining, non-forwards on the team. 9/11 mean 9 (non-forwards) out of 11 total players. The selections are multiplied because you need to consider each possible outcome that satisfies the question.

A good way to approach any problem is to read each sentence in the question and see what information you can pull from it. Let's break down the question:

During the break of a football match, the coach will make 3 substitutions.

3 substitutions will need to be made by the coach. Ok - so we know that the coach wants to make 3 substitutes during the break

If the team consists of 11 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted?

So now we are provided with some information regarding the potential substitutes:
1) There are a total of 11 players
2) Out of the 11 players, 2 are forwards and the rest are not

The last line provides the question that needs to be solved.
what is the probability that none of the forwards will be substituted

Now we know what we need to solve. To summarize: Out of 11 players, 2 are forwards. What is the probability of selecting 3 players, who are not forwards?

First selection - 9 members who are not forwards can be chosen. The group total is 11. So 9 out of 11 can be chosen. 9/11
Second selection - Now you have 8 members left who are not forwards. The total members of the group are 10 (since you selected 1 member previously for the first selection). 8/10
Third selection - Now there are 7 members, who are not forwards that can be selected. The group total is now 9 (because you have chosen 2 people from the previous 2 selections). 7/9

The 3 selections have to be multiplied because each will need to occur in order to calculate the total possibility. You can think of it as selection 1 and selection 2 and selection 3. 9/11 *8/10 * 7/9

The problem can also be solved by using combinatorics.

9C3/11C3

9C3 = 9 Choose 3 = 9!/3!*6! = favorable outcome
11C3 = 11 choose 3 = 11!/3! * 8! = total possibilities

9/11 x 8/10 x 7/9 = 28/55

Answer is C.

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