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Probability question

2016 · Shift 2 · Q23

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JEE Advanced
Mathematics
Probability
2016 · Shift 2 · Q23

Probability question

2016 · Shift 2 · Q23

JEE AdvancedMathematicsProbabilityMCQ+3 / −1

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

{1 \over 2},{1 \over 6}

and

{1 \over 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\,\left( {X = Y} \right)

is

A
${{11} \over {36}}$
B
${{1} \over {3}}$
C
${{13} \over {36}}$
D
${{1} \over {2}}$

View written solutionFree

Correct answer: C

Points in one game

For team $T_1$ against $T_2$ :

Win with probability $\dfrac12$ $\Rightarrow$ points $(X,Y)=(3,0)$
Draw with probability $\dfrac16$ $\Rightarrow$ points $(X,Y)=(1,1)$
Lose with probability $\dfrac13$ $\Rightarrow$ points $(X,Y)=(0,3)$

The two games are independent.

Condition for $X=Y$ after two games

Let us look at the difference in points in one game:

If $T_1$ wins: $X-Y=3$
If draw: $X-Y=0$
If $T_1$ loses: $X-Y=-3$

After two games, for total points to be equal, the total difference must be $0$ .

So the possible game-outcome pairs are:

Draw, Draw
Win, Lose
Lose, Win

No other combination gives equal total points.

Compute probability of each case

Since games are independent,

$P(\text{Draw, Draw})=\dfrac16\cdot\dfrac16=\dfrac1{36}$
$P(\text{Win, Lose})=\dfrac12\cdot\dfrac13=\dfrac16$
$P(\text{Lose, Win})=\dfrac13\cdot\dfrac12=\dfrac16$

Therefore,

P(X=Y)=\frac1{36}+\frac16+\frac16

=\frac1{36}+\frac6{36}+\frac6{36}=\frac{13}{36}

Match with options

P(X=Y)=\frac{13}{36}

Hence the correct option is C.

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