Football teams T1 and T2 have to play...

MathematicsAddition and Multiplication TheoremsJEE Advanced 2016Moderate

Comprehension Passage

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 1/2, 1/6 and 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.

Question 1:

$P (X > Y)$ is

(A)

1/4

(B)

5/12

(C)

1/2

(D)

7/12

Answer:B

Question 2:

$P (X = Y)$ is

(A)

11/36

(B)

1/3

(C)

13/36

(D)

1/2

Answer:C

Visualized Solution (English)

Defining Probabilities

Probabilities for Team $T_{1}$ in a single game:
Win: $P (W) = \frac{1}{2}$ (3 points)
Draw: $P (D) = \frac{1}{6}$ (1 point)
Loss: $P (L) = \frac{1}{3}$ (0 points)

The Sample Space

Total outcomes = $3 \times 3 = 9$
Since games are independent: $P (G_{1} \cap G_{2}) = P (G_{1}) \times P (G_{2})$
We represent outcomes as pairs: $(G am e 1, G am e 2)$

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X > Y : Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W W

Condition: $X > Y$
Case 1: $W W$ (Win, Win)
Points: $T_{1} = 3 + 3 = 6$ , $T_{2} = 0 + 0 = 0$
Probability: $P (W W) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X > Y : Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W D and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D W

Case 2: $W D$ or $D W$ (One Win, One Draw)
Points: $T_{1} = 3 + 1 = 4$ , $T_{2} = 0 + 1 = 1$
Probability: $P (W D) + P (D W) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}$

Total Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (X > Y)

Total $P (X > Y) = P (W W) + P (W D) + P (D W)$
Calculation: $\frac{1}{4} + \frac{1}{12} + \frac{1}{12} = \frac{1}{4} + \frac{1}{6}$
Final Result: $P (X > Y) = \frac{3 + 2}{12} = \frac{5}{12}$

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X = Y : Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W L and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L W

Condition: $X = Y$
Case 1: $W L$ or $L W$ (One Win, One Loss)
Points: $T_{1} = 3 + 0 = 3$ , $T_{2} = 0 + 3 = 3$
Probability: $P (W L) + P (L W) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X = Y : Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D D

Case 2: $D D$ (Two Draws)
Points: $T_{1} = 1 + 1 = 2$ , $T_{2} = 1 + 1 = 2$
Probability: $P (D D) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

Final Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (X = Y)

Total $P (X = Y) = P (W L) + P (L W) + P (D D)$
Calculation: $\frac{1}{3} + \frac{1}{36}$
Final Result: $P (X = Y) = \frac{12 + 1}{36} = \frac{13}{36}$

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X>Y) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=Y) is

Visualized Solution (English)

Defining Probabilities

The Sample Space

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X>Y: Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } WW

Total Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X>Y)

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=Y: Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DD

Final Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=Y)

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