JEE Advanced 2022

LEVELJEE Main

Animated Solution for Mathematics - Probability: Two players, $P_{1}$ and $P_{2}$ , play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let $x$ and $y$ denote the readings on the die rolled by $P_{1}$ and $P_{2}$ , respectively. If $x < y$ , then $P_{1}$ scores 5 points and $P_{2}$ scores 0 point. If $x = y$ , then each player scores 2 points. If $x > y$ , then $P_{1}$ scores 0 point and $P_{2}$ scores 5 points. Let $X_{i}$ and $Y_{i}$ be the total scores of $P_{1}$ and $P_{2}$ , respectively, after playing the $i^{t h}$ round.

Q: Is it possible to have a draw after three rounds if the scores are not equal?

No, the definition of a draw in this game is strictly $X_i = Y_i$. If the scores are not equal, the game is not a draw. The condition $X_3 = Y_3$ requires the total points to be balanced, which only happens if the number of wins and losses are equal, or if all rounds are draws. Keep your focus on the score totals!

List-I

(P)

Probability of

(X_{2} \geq Y_{2})

(Q)

Probability of

(X_{2} > Y_{2})

(R)

Probability of

(X_{3} = Y_{3})

(S)

Probability of

(X_{3} > Y_{3})

List-II

(1)

3/8

(2)

11/16

(3)

5/16

(4)

355/864

(5)

77/432

Select Matching Pairs:

* Multiple Allowed

PMatches

QMatches

RMatches

SMatches

Progressive Hints

Hint 1

Hint 1: The Spark. Before diving into multiple rounds, focus on the single-round probability space. Since the dice are fair, the total outcomes are

6 \times 6 = 36

. Identify the number of outcomes for a draw (

x = y

), and use the symmetry of the dice to determine the probabilities of

P_{1}

winning (

x < y

) and

P_{2}

winning (

x > y

Hint 2

Hint 2: The Arsenal. When dealing with multiple rounds, treat each round as an independent event. For

X_{n} > Y_{n}

X_{n} = Y_{n}

, list the possible sequences of outcomes (Win, Loss, Draw) that lead to these states. Remember that the order of outcomes matters (e.g., Win-Draw is distinct from Draw-Win).

Hint 3

Hint 3: The Execution. Start by calculating the base probabilities:

P (D) = \frac{1}{6}

P (W) = \frac{5}{12}

, and

P (L) = \frac{5}{12}

. For

P (X_{2} > Y_{2})

, sum the probabilities of the favorable sequences:

P (W, W) + P (W, D) + P (D, W)

. Use this logic to build your path for three rounds.

Visualized Solution

$P_{1}$ vs $P_{2}$ : Single Round

Let's define the events for player $P_{1}$ in a single round.
Win ( $W$ ): $x < y ⟹$ Score: $(5, 0)$
Draw ( $D$ ): $x = y ⟹$ Score: $(2, 2)$
Loss ( $L$ ): $x > y ⟹$ Score: $(0, 5)$

Probability of a Draw

Total possible outcomes when rolling two dice $= 36$ .
Number of outcomes where $x = y$ is $6$ (i.e., $(1, 1), (2, 2), \dots, (6, 6)$ ).
$P (D) = \frac{6}{36} = \frac{1}{6}$

Probabilities of Win and Loss

By symmetry, the probability of $x < y$ is equal to $x > y$ .
$P (W) = P (L) = \frac{1 - P ( D )}{2}$
$P (W) = P (L) = \frac{1 - \frac{1}{6}}{2} = \frac{5}{12}$

Two Rounds: $X_{2} > Y_{2}$

We need $P_{1}$ 's score strictly greater than $P_{2}$ 's score after 2 rounds.
Possible favorable sequences for $(P_{1}, P_{2})$ :
$(W, W) ⟹ (10, 0)$
$(W, D)$ or $(D, W) ⟹ (7, 2)$

Computing $P (X_{2} > Y_{2})$

$P (X_{2} > Y_{2}) = P (W, W) + P (W, D) + P (D, W)$
$P (X_{2} > Y_{2}) = P (W)^{2} + 2 \cdot P (W) \cdot P (D)$
$= (\frac{5}{12})^{2} + 2 (\frac{5}{12}) (\frac{1}{6})$
$= \frac{25}{144} + \frac{10}{72} = \frac{45}{144} = \frac{5}{16}$

Two Rounds: $X_{2} = Y_{2}$

Now, let's find the probability of a tie after 2 rounds ( $X_{2} = Y_{2}$ ).
Possible sequences for equal scores:
$(W, L)$ or $(L, W) ⟹ (5, 5)$
$(D, D) ⟹ (4, 4)$

Computing $P (X_{2} = Y_{2})$

$P (X_{2} = Y_{2}) = 2 \cdot P (W) \cdot P (L) + P (D)^{2}$
$= 2 (\frac{5}{12}) (\frac{5}{12}) + (\frac{1}{6})^{2}$
$= \frac{50}{144} + \frac{4}{144} = \frac{54}{144} = \frac{3}{8}$

Computing $P (X_{2} \geq Y_{2})$

The event $X_{2} \geq Y_{2}$ means $P_{1}$ either wins or ties after 2 rounds.
$P (X_{2} \geq Y_{2}) = P (X_{2} > Y_{2}) + P (X_{2} = Y_{2})$
$= \frac{5}{16} + \frac{3}{8}$
$= \frac{5}{16} + \frac{6}{16} = \frac{11}{16}$

Three Rounds: $X_{3} = Y_{3}$

Now consider 3 rounds. When is $X_{3} = Y_{3}$ ?
The total score must be equal. This requires the number of Wins to equal the number of Losses.
Possible sequences:
One $W$ , One $L$ , One $D$ (in any order, $3! = 6$ ways)
Three $D$ 's (all draws, $1$ way)

Computing $P (X_{3} = Y_{3})$

$P (X_{3} = Y_{3}) = 6 \cdot P (W) P (L) P (D) + P (D)^{3}$
$= 6 (\frac{5}{12}) (\frac{5}{12}) (\frac{1}{6}) + (\frac{1}{6})^{3}$
$= \frac{150}{864} + \frac{1}{216}$
$= \frac{150}{864} + \frac{4}{864} = \frac{154}{864} = \frac{77}{432}$

Three Rounds: $X_{3} > Y_{3}$

Finally, we need $P (X_{3} > Y_{3})$ .
By the Law of Total Probability:
$P (X_{3} > Y_{3}) + P (X_{3} = Y_{3}) + P (X_{3} < Y_{3}) = 1$
Due to the symmetric nature of the game, $P_{1}$ winning is just as likely as $P_{2}$ winning:
$P (X_{3} > Y_{3}) = P (X_{3} < Y_{3})$

Computing $P (X_{3} > Y_{3})$

Let $P (X_{3} > Y_{3}) = p$ .
$2 p + P (X_{3} = Y_{3}) = 1$
$2 p + \frac{77}{432} = 1 ⟹ 2 p = 1 - \frac{77}{432} = \frac{355}{432}$
$p = \frac{355}{864}$

Final Conclusion

(I) $P (X_{2} \geq Y_{2}) = \frac{11}{16}$ $\to$ Option 2
(II) $P (X_{2} > Y_{2}) = \frac{5}{16}$ $\to$ Option 3
(III) $P (X_{3} = Y_{3}) = \frac{77}{432}$ $\to$ Option 5
(IV) $P (X_{3} > Y_{3}) = \frac{355}{864}$ $\to$ Option 4
The correct matching is I $\to$ 2, II $\to$ 3, III $\to$ 5, IV $\to$ 4.

00:00 / 00:00

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Single Round Foundation

Before we can conquer three rounds, we must understand the heartbeat of a single round. We have two players,

P_{1}

and

P_{2}

, each rolling a fair six-sided die. The total number of outcomes is

6 \times 6 = 36

A draw occurs when

x = y

. There are exactly 6 such pairs:

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

. Thus, the probability of a draw is:

P (D) = \frac{6}{36} = \frac{1}{6}

By the sheer beauty of symmetry, the probability of

P_{1}

winning (

x < y

) must be identical to the probability of

P_{1}

losing (

x > y

). Since the total probability must sum to 1, we have:

P (W) = P (L) = \frac{1 - P ( D )}{2} = \frac{1 - 1/6}{2} = \frac{5}{12}

The Two-Round Complexity

Now, the game intensifies. We play two rounds. We want to know the probability that

P_{1}

is leading (

X_{2} > Y_{2}

P_{1}

could win both rounds, or win one and draw one. We map the sequences:

(W, W)

(W, D)

, or

(D, W)

The probability of

(W, W)

P (W)^{2} = (5/12)^{2} = 25/144

. The probability of

(W, D)

P (W) \times P (D) = (5/12) \times (1/6) = 5/72

Summing these probabilities, we get:

\frac{25}{144} + \frac{10}{72} = \frac{25}{144} + \frac{20}{144} = \frac{45}{144} = \frac{5}{16}

The same logic applies to the tie condition

X_{2} = Y_{2}

. We could have

(W, L)

(L, W)

, or

(D, D)

. Calculating this gives:

2 \times (\frac{5}{12} \times \frac{5}{12}) + (\frac{1}{6})^{2} = \frac{50}{144} + \frac{4}{144} = \frac{54}{144} = \frac{3}{8}

The Three-Round Symmetry

Finally, we reach the three-round challenge. For

X_{3} = Y_{3}

, the scores must balance. This happens if we have one win, one loss, and one draw (in any of the

3! = 6

permutations) or if all three rounds are draws.

The probability is:

6 \times P (W) P (L) P (D) + P (D)^{3} = 6 \times (\frac{5}{12}) \times (\frac{5}{12}) \times (\frac{1}{6}) + (\frac{1}{6})^{3}

= \frac{150}{864} + \frac{4}{864} = \frac{154}{864} = \frac{77}{432}

Now, for

P (X_{3} > Y_{3})

, we use the ultimate JEE weapon: Symmetry. We know that

P (X_{3} > Y_{3}) + P (X_{3} = Y_{3}) + P (X_{3} < Y_{3}) = 1

Since

P (X_{3} > Y_{3}) = P (X_{3} < Y_{3})

, let this be

p

. Then:

2 p + \frac{77}{432} = 1

Solving for

p

, we get

2 p = 355/432

, which yields the final result:

p = \frac{355}{864}

Common Traps & FAQs

Wait, why is the probability of winning equal to the probability of losing? Shouldn't the dice rolls be different?

That is a fantastic observation! Because both players are rolling identical, fair dice, the game is perfectly symmetric. For every outcome where

P_{1}

rolls a 1 and

P_{2}

rolls a 2, there is a corresponding outcome where

P_{1}

rolls a 2 and

P_{2}

rolls a 1. Since the dice are fair, these outcomes are equally likely. Thus,

P (x < y)

must equal

P (x > y)

Why do we multiply by 2 when calculating the probability of a Win-Draw sequence?

Think of it as the order of events. A 'Win' followed by a 'Draw' is a distinct sequence from a 'Draw' followed by a 'Win'. Since both sequences result in the same final state, we must account for both paths in our probability sum. It is a classic counting principle: if there are

n

ways to achieve a state, we must sum the probabilities of all those distinct paths.

Why can't we just use energy conservation or some other shortcut for the three-round case?

In probability, we use the Law of Total Probability and symmetry as our 'shortcuts'. For three rounds, calculating every possible sequence is tedious. Instead, we recognize that

P (X_{3} > Y_{3}) + P (X_{3} = Y_{3}) + P (X_{3} < Y_{3}) = 1

. By symmetry,

P (X_{3} > Y_{3}) = P (X_{3} < Y_{3})

. This turns a complex counting problem into a simple algebraic equation:

2 p + P (X_{3} = Y_{3}) = 1

Is it possible to have a draw after three rounds if the scores are not equal?

No, the definition of a draw in this game is strictly

X_{i} = Y_{i}

. If the scores are not equal, the game is not a draw. The condition

X_{3} = Y_{3}

requires the total points to be balanced, which only happens if the number of wins and losses are equal, or if all rounds are draws. Keep your focus on the score totals!

List-I

List-II

Select Matching Pairs:

Visualized Solution

$P_{1}$ vs $P_{2}$ : Single Round

Probability of a Draw

Probabilities of Win and Loss

Two Rounds: $X_{2} > Y_{2}$

Computing $P (X_{2} > Y_{2})$

Two Rounds: $X_{2} = Y_{2}$

Computing $P (X_{2} = Y_{2})$

Computing $P (X_{2} \geq Y_{2})$

Three Rounds: $X_{3} = Y_{3}$

Computing $P (X_{3} = Y_{3})$

Three Rounds: $X_{3} > Y_{3}$

Computing $P (X_{3} > Y_{3})$

Final Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Single Round Foundation

The Two-Round Complexity

The Three-Round Symmetry

Similar Questions

Comprehension Passage

$P (X > Y)$ is

$P (X = Y)$ is

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is

India plays two matches each with West Indies and Australia. In any match the probabilities of India getting, points $0, 1$ and $2$ are $0.45, 0.05$ and $0.50$ respectively. Assuming that the outcomes are independent, the probability of India getting at least $7$ points is

Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If $E_{1}$ is the event that die $A$ shows up four, $E_{2}$ is the event that die $B$ shows up two and $E_{3}$ is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true?

An unbiased die with faces marked $1, 2, 3, 4, 5$ and $6$ is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than $2$ and the maximum face value is not greater than $5$ , is

Three faces of a fair die are yellow, two faces red and one blue. The die is tossed three times. The probability that the colours, yellow, red and blue, appear in the first, second and the third tosses respectively is .........

For two given events $A$ and $B, P (A \cap B)$ is

List-I

List-II

Select Matching Pairs:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​ vs .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​: Single Round

Probability of a Draw

Probabilities of Win and Loss

Two Rounds: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X2​>Y2​

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X2​>Y2​)

Two Rounds: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X2​=Y2​

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X2​=Y2​)

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X2​≥Y2​)

Three Rounds: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X3​=Y3​

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X3​=Y3​)

Three Rounds: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X3​>Y3​

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X3​>Y3​)

Final Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Single Round Foundation

The Two-Round Complexity

The Three-Round Symmetry

Similar Questions

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

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is

Three faces of a fair die are yellow, two faces red and one blue. The die is tossed three times. The probability that the colours, yellow, red and blue, appear in the first, second and the third tosses respectively is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For two given events A and B,P(A∩B) is

$P_{1}$ vs $P_{2}$ : Single Round

Two Rounds: $X_{2} > Y_{2}$

Computing $P (X_{2} > Y_{2})$

Two Rounds: $X_{2} = Y_{2}$

Computing $P (X_{2} = Y_{2})$

Computing $P (X_{2} \geq Y_{2})$

Three Rounds: $X_{3} = Y_{3}$

Computing $P (X_{3} = Y_{3})$

Three Rounds: $X_{3} > Y_{3}$

Computing $P (X_{3} > Y_{3})$

$P (X > Y)$ is

$P (X = Y)$ is

For two given events $A$ and $B, P (A \cap B)$ is