JEE Main 2019 (12 January)

LEVELJEE Main

Animated Solution for Mathematics - Probability: In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is :

Select Answer:

Visualized Solution

Defining Probabilities

Winning means getting a $5$ or $6$ .
$P (W) = \frac{2}{6} = \frac{1}{3}$
Losing means getting $1, 2, 3,$ or $4$ .
$P (L) = \frac{4}{6} = \frac{2}{3}$

Stopping Conditions

The game stops if the man wins.
The game also stops after a maximum of $3$ throws.

Case 1: Win on 1st Throw

Case 1: Win on the first throw ( $W$ ).
Net Gain: $+ 100$
Probability: $P (W) = \frac{1}{3}$

First Throw Loss

What if he loses the first throw?
Loss on 1st throw ( $L$ ).
Penalty: $- 50$

Case 2: Loss then Win

Case 2: Loss on 1st, Win on 2nd ( $L, W$ ).
Net Gain: $- 50 + 100 = + 50$
Probability: $P (L) \times P (W) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$

Second Throw Loss

What if he loses the second throw too?
Loss on 1st, Loss on 2nd ( $L, L$ ).
Penalty so far: $- 50 - 50 = - 100$

Case 3: Two Losses then Win

Case 3: Loss, Loss, Win ( $L, L, W$ ).
Net Gain: $- 50 - 50 + 100 = 0$
Probability: $P (L) \times P (L) \times P (W) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27}$

Case 4: Three Consecutive Losses

Case 4: Loss, Loss, Loss ( $L, L, L$ ).
Net Gain: $- 50 - 50 - 50 = - 150$
Probability: $P (L) \times P (L) \times P (L) = (\frac{2}{3})^{3} = \frac{8}{27}$

Expected Value Formula

Expected Value $E (X) = \sum X_{i} P (X_{i})$
$X_{i}$ = Net Gain in each case
$P (X_{i})$ = Probability of that case

Substituting the Values

$E (X) = (100) (\frac{1}{3}) + (50) (\frac{2}{9}) + (0) (\frac{4}{27}) + (- 150) (\frac{8}{27})$

Calculating the Expected Value

$E (X) = \frac{100}{3} + \frac{100}{9} + 0 - \frac{1200}{27}$
Common denominator is $27$ :
$E (X) = \frac{900}{27} + \frac{300}{27} - \frac{1200}{27}$

Final Result

$E (X) = \frac{900 + 300 - 1200}{27}$
$E (X) = \frac{1200 - 1200}{27} = 0$
The expected gain/loss is Rs. $0$ .

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The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Analyzing the Setup

To solve this problem, we must visualize the game as a branching path, or a probability tree. On any single throw, the probability of winning,

P (W)

, is

\frac{2}{6}

, which simplifies to

P (W) = \frac{1}{3}

Conversely, the probability of losing,

P (L)

, is

\frac{4}{6}

, which simplifies to

P (L) = \frac{2}{3}

. We are limited to a maximum of three throws, and the game terminates immediately upon a win.

Mapping the Possibilities

We define the outcomes based on the sequence of events:

Path 1: You win on the first throw. The probability is

\frac{1}{3}

, and your gain is

+ 100

Path 2: You lose the first, then win the second. The probability is

P (L) \times P (W) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}

. Your net gain is

- 50 + 100 = + 50

Path 3: You lose the first, lose the second, then win the third. The probability is

P (L) \times P (L) \times P (W) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27}

. Your net gain is

- 50 - 50 + 100 = 0

Path 4: You lose all three throws. The probability is

P (L) \times P (L) \times P (L) = (\frac{2}{3})^{3} = \frac{8}{27}

. Your net gain is

- 50 - 50 - 50 = - 150

The Expected Value Synthesis

The Expected Value,

E (X)

, is the weighted average of all possible outcomes, defined by the formula

E (X) = \sum x_{i} p_{i}

. Substituting our calculated values, we get:

E (X) = (100) (\frac{1}{3}) + (50) (\frac{2}{9}) + (0) (\frac{4}{27}) + (- 150) (\frac{8}{27})

To solve this, we find a common denominator of

27

E (X) = \frac{900}{27} + \frac{300}{27} + 0 - \frac{1200}{27}

Final Calculation

Combining the terms in the numerator:

E (X) = \frac{900 + 300 - 1200}{27} = \frac{1200 - 1200}{27} = 0

The result is 0. This means that, on average, you neither win nor lose money. In the world of probability, this is defined as a fair game.

Common Traps & FAQs

Why do we stop at three throws? Does the rule imply he loses if he doesn't win by then?

The rule states he plays until he gets a 5 or 6 OR until he reaches a maximum of three throws. This means the game ends regardless of the outcome after the third throw. If he hasn't won by the third throw, he simply stops playing, and his net gain is calculated based on the losses accumulated up to that point.

In the 'Lose-Lose-Win' case, why is the net gain zero?

Let's break it down: he loses Rs. 50 on the first throw, another Rs. 50 on the second throw, and then wins Rs. 100 on the third throw. The net gain is

- 50 - 50 + 100 = 0

. He essentially breaks even, which is a perfectly valid outcome in probability games!

Is there a faster way to solve this without drawing a tree?

While a tree is the most intuitive way to visualize the paths, you could also use a recursive approach or a simple summation. However, for a small number of throws like three, the tree method is the most robust way to ensure you don't miss any scenarios or miscalculate the probabilities of the sequences.

Select Answer:

Visualized Solution

Defining Probabilities

Stopping Conditions

Case 1: Win on 1st Throw

First Throw Loss

Case 2: Loss then Win

Second Throw Loss

Case 3: Two Losses then Win

Case 4: Three Consecutive Losses

Expected Value Formula

Substituting the Values

Calculating the Expected Value

Final Result

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

Mapping the Possibilities

The Expected Value Synthesis

Final Calculation

Similar Questions

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Comprehension Passage

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Select Answer:

Visualized Solution

Defining Probabilities

Stopping Conditions

Case 1: Win on 1st Throw

First Throw Loss

Case 2: Loss then Win

Second Throw Loss

Case 3: Two Losses then Win

Case 4: Three Consecutive Losses

Expected Value Formula

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Calculating the Expected Value

Final Result

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Analyzing the Setup

Mapping the Possibilities

The Expected Value Synthesis

Final Calculation

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