Question

In a certain game A’s skill is to B’s as 3 to 2. Find the chances of A winning 3 games at least out of 5.

Answer 1

Hint: In this question it is given that A’s skill is to B’s as 3 to 2, this means that A’s chances of winning a game is $\dfrac{3}{5}$ using the basic concept of probability that ${\text{Probability = }}\dfrac{{{\text{Favorable chances}}}}{{{\text{Total chances}}}}$, similar we can find B’s chances of winning a game. Use this concept along with the binomial theorem of probability to reach the answer.

Complete step-by-step answer:
It is given that in a certain game A’s skill is to B’s as 3 to 2……………….. (1)
Using ${\text{Probability = }}\dfrac{{{\text{Favorable chances}}}}{{{\text{Total chances}}}}$……………….. (2)
Let p be the probability of A winning a game.

So, the chance of winning by A (i.e. probability to win by A) using equation (1) and equation (2)
$ \Rightarrow p = \dfrac{3}{{3 + 2}} = \dfrac{3}{5}$.

Let q be the probability of B winning a game. So the probability to win by B using equation (1) and equation (2)
$ \Rightarrow q = \dfrac{2}{{3 + 2}} = \dfrac{2}{5}$

Now we have to find out the chance of A winning 3 games at least out of 5.

Now we use binomial theorem to calculate the required probability.

$ \Rightarrow {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}$
Where, n = number of times the game played.
               r = number of times A wins.
               p = Probability to win by A.
               q = Probability to win by B, or losing probability of A.

Now it is given A win at least 3 games (i.e. A win 3 games or 4 games or 5 games)
So, the required probability (${P_A}$) of winning 3 games at least out of 5 is
${P_A} = {}^5{C_3}{\left( {\dfrac{3}{5}} \right)^3}{\left( {\dfrac{2}{5}} \right)^{5 - 3}} + {}^5{C_4}{\left( {\dfrac{3}{5}} \right)^4}{\left( {\dfrac{2}{5}} \right)^{5 - 4}} + {}^5{C_5}{\left( {\dfrac{3}{5}} \right)^5}{\left( {\dfrac{2}{5}} \right)^{5 - 5}}$…………… (3)
Now as we know ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$
   \Rightarrow {}^5{C_3} = \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} = \dfrac{{5.4.3!}}{{3!\left( {2 \times 1} \right)}} = 10 \\
   \Rightarrow {}^5{C_4} = \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} = \dfrac{{5.4!}}{{4!\left( 1 \right)}} = 5 \\
   \Rightarrow {}^5{C_5} = \dfrac{{5!}}{{5!\left( {5 - 5} \right)!}} = \dfrac{{5!}}{{5!\left( {0!} \right)}} = \dfrac{{5!}}{{5!\left( 1 \right)}} = 1 \\
 $
So substitute these values in equation (3) we have

${P_A} = 10{\left( {\dfrac{3}{5}} \right)^3}{\left( {\dfrac{2}{5}} \right)^2} + 5{\left( {\dfrac{3}{5}} \right)^4}{\left( {\dfrac{2}{5}} \right)^1} + 1{\left( {\dfrac{3}{5}} \right)^5}{\left( {\dfrac{2}{5}} \right)^0}$
Now simplify the above equation we have,
${P_A} = 10\left( {\dfrac{{27}}{{125}}} \right)\left( {\dfrac{4}{{25}}} \right) + 5\left( {\dfrac{{81}}{{625}}} \right)\left( {\dfrac{2}{5}} \right) + 1\left( {\dfrac{{243}}{{3125}}} \right)\left( 1 \right)$
${P_A} = \dfrac{{1080}}{{3125}} + \dfrac{{810}}{{3125}} + \dfrac{{243}}{{3125}}$
Now add all these terms we have,
${P_A} = \dfrac{{2133}}{{3125}}$

So, this is the required probability of A winning 3 games at least out of 5.
So, this is the required answer.

Note: Whenever we face such types of problems the key point is to take out individual probabilities of winning a specific game then whenever we come across words in questions like at least or at most involving probability then we always have to use the binomial probability theorem. This will help you get on the right track to reach the answer.