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Suppose in a competition $11$ matches are to be played, each having one of $3$ distinct outcomes as possibilities. The number of ways one can predict the outcomes of all $11$ matches such that exactly $6$ of the predictions turn out to be correct is

1. $\begin{pmatrix}11 \\ 6 \end{pmatrix} \times 2^5$
2. $\begin{pmatrix}11 \\ 6 \end{pmatrix}$
3. $3^6$
4. none of the above

recategorized | 36 views

Consider the outcomes of each match being win(w) , loss(l) or draw(d)...

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Now for 6matches the prediction turns out to be correct which can be done in 11C6 ways but now other five matches will have incorrect prediction means if we predict w it should be l or d.. and soon..

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Therefore for five matches will be predicted in  2^5 ways...

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Hence total number of predictions will be 11C6* 2^5. Option A...
by Boss (12.1k points)
Each match can have outcome Win, Loose or Tie, so total possibilities of outcome  $= 3^{11}$

But we have to find exactly 6 out of any 11 matches, so this can be done in $11 \choose 6$ ways.

Now for each of the chosen match must have correct prediction which can be done in only $1$ way.

So the total number of ways of predicting exactly 6 matches correctly

= $11 \choose 6$ $* 1$

= $11 \choose 6$

Hence answer is $B$
by Active (2.2k points)
0
Can u recheck the solution :)I think you have missed something.
0
Correct match prediction can be done only in one way

Not getting what i have missed

Each match has 3 possible outcomes.

For 11 matches,on total there are $3^{11}$ ways of prediction.

for each match,probability that our prediction is correct is 1/3.

Hence the probability that our prediction is wrong is 2/3.

Let X be the random variable that counts number of correct predictions.

therefore P(X=6)=$\binom{11}{6}$*$\left ( 1/3 \right )^{6}$*$\left ( 2/3 \right )^{5}$.(by Binomial theorem)

Total number of ways in which exactly 6 predictions are correct= probability that exactly 6 predictions are correct*Total number of predictions.

=$\binom{11}{6}$*$\left ( 1/3 \right )^{6}$*$\left ( 2/3 \right )^{5}$  * $3^{11}$.

=$\binom{11}{6}$ * $2^{5}$

by Active (3.2k points)