Question

Football teams ${T_1}$​ and ${T_2}$​ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of ${T_1}$ winning, drawing and losing a game against ${T_2}$​ are $\dfrac{1}{2}$ ​, $\dfrac{1}{6}$ and $\dfrac{1}{3}$ respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams ${T_1}$​ and ${T_2}$ respectively, after two games. P(X=Y) is:
 $
  A.\dfrac{{11}}{{36}} \\
  B.\dfrac{1}{3} \\
  C.\dfrac{{13}}{{36}} \\
  D.\dfrac{1}{{12}} \\
 $

Answer 1

Hint: This question is a good example of probability consisting of independent events. First identify the possible outcomes for ${T_1}$ and ${T_2}$ with respect to the cases of win, loss and draw. Then show each with their probabilities. Identify three possible ways in which (X = Y) after two games. Add the probabilities of all cases as identified. This will be the result probability.

Complete step-by-step answer:
It is given in the question that,
Points given for win= 3
Points given for loss= 0
Points given for draw= 1
Also the probabilities for ${T_1}$ against ${T_2}$ ,
For winning is $\dfrac{1}{2}$ . So, $P[{T_1}] = \dfrac{1}{2} $….(1)
For losing is $\dfrac{1}{3}$ (Means winning of ${T_2}$) .So, $P[{T_2}] = \dfrac{1}{3}$ ….(2)
For draw is $\dfrac{1}{6}$ . So, $P[D] = \dfrac{1}{6}$ ….(3)
It is given that X and Y denote the total points scored by teams ${T_1}$​ and ${T_2}$ respectively,
Now, we have to obtain the events which will be favourable for X = Y after two games.
For X = Y , three favourable events will be:
${T_1}$ win , ${T_2}$ win
Draw and draw
${T_2}$ win and ${T_1}$ win.
Thus,
P(X=Y) = P[ $ {T_1} $$ {T_2}$ ]+P[D D]+P[ ${T_2}$ ${T_1}$ ] , where D represents draw.
Here we have added all probabilities, as one of the three will occur. But in individual case we need to multiply the probabilities, as both will occur together.
Now, P(X=Y) = P[ ${T_1}$${T_2}$ ]+P[D D]+P[${T_2}$${T_1}$ ]
\[
   \Rightarrow P\left( {X = Y} \right) = P[{T_1}{T_2}\left] { + P} \right[D{D_1}\left] { + P} \right[{T_2}{T_1}] \\
   \Rightarrow P\left( {X = Y} \right) = P[{T_1}] \times P[{T_2}] + P[D] \times P[D] + P[{T_2}] \times P[{T_1}] \\
    \\
 \]
Substituting the values from equations (1) and (3) in above equation, we get:
 \[
  P\left( {X = Y} \right) = \dfrac{1}{2} \times \dfrac{1}{3} + \dfrac{1}{6} \times \dfrac{1}{6} + \dfrac{1}{3} \times \dfrac{1}{2} \\
  P\left( {X = Y} \right) = \dfrac{1}{6} + \dfrac{1}{{36}} + \dfrac{1}{6} \\
  P\left( {X = Y} \right) = \dfrac{{13}}{{36}} \\
    \\
 \]
$ \therefore $Probability of P(X=Y) will be $\dfrac{{13}}{{36}}$ .

Thus option C is correct.

Note: Probability is a very interesting and practical oriented branch of mathematics. It involves the occurrences and the chances of events with many possibilities. In this question we have used both additive and multiplicative principles of probability. Also, with careful understanding of the question, its solution will be straightforward.