Question

A contest consists of predicting the results of win, draw or defeat of 7 football matches. A sent his entry by predicting at random. The probability that his entry will contain exactly 4 correct predictions is
A). \[\dfrac{8}{{{3^7}}}\]
B). \[\dfrac{{16}}{{{3^7}}}\]
C). \[\dfrac{{280}}{{{3^7}}}\]
D). \[\dfrac{{560}}{{{3^7}}}\]

Answer 1

Hint: The given event has three possibilities actually present that is results of win, draw or defeat. But we will deal with only one successful case of any one of the result. Thus we will get the probability of successful and failed event. Then we will use bernauli’s principle. This involves the probability of the events that have successful and failure of attempts.

Complete step-by-step solution:
Given that the total number of matches is 7.
But the probability of any successful event is \[\dfrac{1}{3}\]
And that of failed event is \[\dfrac{2}{3}\]
Now let p= \[\dfrac{1}{3}\] and q=\[\dfrac{2}{3}\]
As we know the successful attempts are given to be 4.
Thus we can use the formula as,
\[P\left( {x = r} \right) = \sum {{}^n{C_r}} {\left( p \right)^r}{\left( q \right)^{n - r}}\]
Total events are n=7, successful events r=4
\[P\left( {x = 4} \right) = \sum {{}^7{C_4}} {\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^{7 - 4}}\]
\[P\left( {x = 4} \right) = {}^7{C_4}{\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^3}\]
\[P\left( {x = 4} \right) = \dfrac{{7!}}{{4!3!}}{\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^3}\]
Factorial can be solved as,
\[P\left( {x = 4} \right) = \dfrac{{7 \times 6 \times 5 \times 4!}}{{4! \times 3 \times 2 \times 1}} \times \dfrac{1}{{{3^4}}} \times \dfrac{{{2^3}}}{{{3^3}}}\]
The denominators are with same base thus the powers can be added,
\[P\left( {x = 4} \right) = 7 \times 5 \times \dfrac{{{2^3}}}{{{3^7}}}\]
Third power of 2 is 8.
\[P\left( {x = 4} \right) = \dfrac{{7 \times 5 \times 8}}{{{3^7}}}\]
On solving we get,
\[P\left( {x = 4} \right) = \dfrac{{280}}{{{3^7}}}\]
This is the correct answer.
Thus option C is the correct option.

Note: Note that 4 predictions are definitely given to be correct. So we took r=4 directly. Also note that whatever the prediction can be, that is either the match is a draw, win or defeat. It is not necessary that every time it is the same result and also not necessary that every time it is different. We don’t know the case ,but we are given that 4 predictions are correct.
Also note that the power of 3 is 7 given in the options already. So no need to calculate the power.