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Probability question

2016 · Shift 2 · Q24

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JEE Advanced
Mathematics
Probability
2016 · Shift 2 · Q24

Probability question

2016 · Shift 2 · Q24

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 \gt Y} \right)

is

A
${1 \over 4}$
B
${5 \over 12}$
C
${1 \over 2}$
D
${7 \over 12}$

View written solutionFree

Correct answer: B

Points in one game

Let us write the points of $(T_1,T_2)$ in one game:

If $T_1$ wins: probability $\dfrac12$ , points $(3,0)$
If draw: probability $\dfrac16$ , points $(1,1)$
If $T_1$ loses: probability $\dfrac13$ , points $(0,3)$

The two games are independent.

We need $P(X>Y),$ where $X,Y$ are total points after two games.

Compare total points via point difference

Define the difference in one game as $D = (\text{points of }T_1) - (\text{points of }T_2).$ Then for one game:

Win $\Rightarrow D=3$ with probability $\dfrac12$
Draw $\Rightarrow D=0$ with probability $\dfrac16$
Loss $\Rightarrow D=-3$ with probability $\dfrac13$

After two games, let total difference be $D_{\text{total}} = X-Y.$ We want $P(X>Y)=P(D_{\text{total}}>0).$

Since each game contributes $3,0,-3$ , divide by $3$ and consider values $1,0,-1$ instead. Let per-game reduced difference be $Z$ :

$Z=1$ with probability $\dfrac12$
$Z=0$ with probability $\dfrac16$
$Z=-1$ with probability $\dfrac13$

For two games, we need $P(Z_1+Z_2>0).$

List favorable cases

Possible pairs $(Z_1,Z_2)$ giving positive sum:

$(1,1)$ gives sum $2$
$(1,0)$ gives sum $1$
$(0,1)$ gives sum $1$

Cases like $(1,-1)$ or $(-1,1)$ give $0$ , so they are not included.

Thus, $P(X>Y)=P(1,1)+P(1,0)+P(0,1).$

Using independence,

$P(1,1)=\frac12\cdot\frac12=\frac14,$ $P(1,0)=\frac12\cdot\frac16=\frac{1}{12},$ $P(0,1)=\frac16\cdot\frac12=\frac{1}{12}.$

So,

Now, $\frac14=\frac{3}{12},$ therefore $P(X>Y)=\frac{3}{12}+\frac{1}{12}+\frac{1}{12}=\frac{5}{12}.$

Check with options

$\frac{5}{12}$ corresponds to Option B.

Comparison with stored answer

Stored correct answer: B

Our derived answer is also B.

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