JEE Advanced 2020

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Animated Solution for Mathematics - Probability: Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If $p$ is the probability that this perfect square is an odd number, then the value of $14 p$ is ____.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the sample space of 36 outcomes. The experiment is a repeated trial, but notice that we only care about the *relative* likelihood of the game ending in a prime sum versus a perfect square sum. The outcomes that do not result in either are effectively 'noise' that we can ignore.

Hint 2

Hint 2 (The Arsenal): This is a classic conditional probability problem. Let

B

be the event that the game ends with a perfect square, and

C

be the event that the game ends with an odd perfect square. Since the game *must* end with either a prime or a perfect square, and we are given that it ended with a perfect square, we are looking for the probability

P (C ∣ B)

Hint 3

Hint 3 (The Execution): The probability

p

is simply the ratio of the number of ways to get an odd perfect square to the total number of ways to get *any* perfect square. Start by calculating

n (B)

(total ways to get a perfect square) and

n (C)

(ways to get an odd perfect square), then use the formula

p = \frac{n ( C )}{n ( B )}

Visualized Solution

Visualizing the Sample Space $S$

Let $S$ be the sum of numbers on two fair dice.
Possible values: $S \in {2, 3, 4, \dots, 12}$
Total number of outcomes $= 6 \times 6 = 36$

The Stopping Condition

The experiment stops when $S$ is a Prime Number or a Perfect Square.
Outcomes that do not stop the game are ignored.

Event $A$ : Prime Sums

Let Event $A$ = The sum is a prime number.
Prime sums: $S \in {2, 3, 5, 7, 11}$

Calculating $P (A)$

Number of ways to get prime sums:
$n (A) = 1 + 2 + 4 + 6 + 2 = 15$
$P (A) = \frac{15}{36}$

Event $B$ : Perfect Square Sums

Let Event $B$ = The sum is a perfect square.
Perfect square sums: $S \in {4, 9}$

Calculating $P (B)$

Number of ways to get perfect square sums:
$n (B) = 3 + 4 = 7$
$P (B) = \frac{7}{36}$

Event $C$ : Odd Perfect Square Sums

Let Event $C$ = The sum is an odd perfect square.
Odd perfect square sums: $S \in {9}$

Calculating $P (C)$

$n (C) = 4 ⟹ P (C) = \frac{4}{36}$

Conditional Probability Logic

Given: A perfect square occurred before a prime.
This means the game ended because Event $B$ occurred.
We need the probability of Event $C$ , given Event $B$ has occurred.
Required Probability $p = P (C ∣ B)$

Applying Conditional Formula

Since $C$ is a subset of $B$ (all odd perfect squares are perfect squares):
$P (C ∣ B) = \frac{P ( C \cap B )}{P ( B )} = \frac{P ( C )}{P ( B )}$
$p = \frac{\frac{4}{36}}{\frac{7}{36}}$

Calculating $p$

Canceling the denominators:
$p = \frac{4}{7}$

Final Calculation for $14 p$

We need to find the value of $14 p$ .
$14 p = 14 \times \frac{4}{7} = 8$
Key Takeaway: In repeated trials, the probability of Event $X$ occurring before Event $Y$ is simply $\frac{P ( X )}{P ( X ) + P ( Y )}$ .

00:00 / 00:00

The Sigma Insight: Conditional Probability

Solution Diagram

Analyzing the Sample Space

When rolling two fair dice, each with faces numbered

1

through

6

, the total number of possible outcomes is

6 \times 6 = 36

. The sum

S

of the two dice can range from

2

(which is

1 + 1

) to

12

(which is

6 + 6

Our sample space

S

is the set

{2, 3, 4, \dots, 12}

Identifying the Stopping Conditions

The game ends when the sum hits a prime number or a perfect square. Let us define these sets:

The prime sums are

A = {2, 3, 5, 7, 11}

The perfect square sums are

B = {4, 9}

Note that these sets are disjoint; no number can be both prime and a perfect square. Any sum that is not in

A \cup B

results in a reroll.

The Conditional Insight

The problem asks for the probability

p

that the perfect square is odd, given that the sum turned out to be a perfect square before it turned out to be a prime. Because the rerolls do not affect the relative probability of hitting one target over the other, we can restrict our sample space to the set of outcomes

A \cup B

We are given that the game ended in a perfect square. This restricts our focus to the set

B = {4, 9}

. Within this set, we seek the probability that the sum is an odd perfect square, which is the subset

C = {9}

The Calculation

Let us count the number of ways to achieve these sums:

For a sum of

4

, the outcomes are

(1, 3), (2, 2), (3, 1)

, which gives

3

ways.

For a sum of

9

, the outcomes are

(3, 6), (4, 5), (5, 4), (6, 3)

, which gives

4

ways.

The total number of ways to get a perfect square is:

n (B) = 3 + 4 = 7

The number of ways to get an odd perfect square (the sum of

9

) is:

n (C) = 4

The conditional probability

p

is the ratio of the favorable outcomes to the total outcomes in our restricted sample space:

p = P (C ∣ B) = \frac{n ( C )}{n ( B )} = \frac{4}{7}

Final Calculation

The question asks for the value of

14 p

. Substituting our value of

p

14 \times \frac{4}{7} = 2 \times 4 = 8

By focusing on the conditional nature of the event rather than the infinite process, we have arrived at the final answer: 8.

Common Traps & FAQs

Wait, do I need to use an infinite geometric series for the repeated rolls?

That is a common instinct! While you *could* model this as an infinite series where the game continues until a stopping condition is met, it is unnecessary. Because the stopping conditions (prime or square) are mutually exclusive in each roll, the probability of the game ending in a specific way is simply the ratio of the probability of that event to the sum of the probabilities of all stopping events. The 'repeated' nature cancels out.

Why is the probability of the prime numbers irrelevant in the final calculation?

Great observation! The question asks for the probability *given* that the sum is a perfect square. This condition restricts our sample space entirely to the set of perfect squares. Once we are 'inside' the world where the game ended in a perfect square, the prime numbers no longer influence the outcome.

Is it possible for a sum to be both a prime and a perfect square?

No, it is not. A prime number has exactly two factors (1 and itself), while a perfect square (greater than 1) has an odd number of factors, including at least one factor other than 1 and itself. Therefore, the sets of prime sums and perfect square sums are disjoint, which simplifies our math significantly.

Enter Numerical Value:

Visualized Solution

Visualizing the Sample Space $S$

The Stopping Condition

Event $A$ : Prime Sums

Calculating $P (A)$

Event $B$ : Perfect Square Sums

Calculating $P (B)$

Event $C$ : Odd Perfect Square Sums

Calculating $P (C)$

Conditional Probability Logic

Applying Conditional Formula

Calculating $p$

Final Calculation for $14 p$

The Sigma Insight: Conditional Probability

Analyzing the Sample Space

Identifying the Stopping Conditions

The Conditional Insight

The Calculation

Final Calculation

Similar Questions

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A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. Then the probability that 5 comes before 7 is .........

Two integers are selected at random from the set ${1, 2, \dots, 11}$ . Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

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If $\overset{ˉ}{E}$ and $\overset{ˉ}{F}$ are the complementary events of events $E$ and $F$ respectively and if $0 < P (F) < 1$ , then

Enter Numerical Value:

Visualized Solution

Visualizing the Sample Space .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

The Stopping Condition

Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A: Prime Sums

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

Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B: Perfect Square Sums

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

Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C: Odd Perfect Square Sums

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

Conditional Probability Logic

Applying Conditional Formula

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 14p

The Sigma Insight: Conditional Probability

Analyzing the Sample Space

Identifying the Stopping Conditions

The Conditional Insight

The Calculation

Final Calculation

Similar Questions

A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two integers are selected at random from the set {1,2,…,11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A and B are two events such that P(A)=31​, P(B)=51​ and P(A∪B)=21​, then P(A∣B′)+P(B∣A′) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let E1​ and E2​ be two events such that the conditional probabilities P(E1​∣E2​)=21​, P(E2​∣E1​)=43​ and P(E1​∩E2​)=81​. Then:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If Eˉ and Fˉ are the complementary events of events E and F respectively and if 0<P(F)<1, then

Visualizing the Sample Space $S$

Event $A$ : Prime Sums

Calculating $P (A)$

Event $B$ : Perfect Square Sums

Calculating $P (B)$

Event $C$ : Odd Perfect Square Sums

Calculating $P (C)$

Calculating $p$

Final Calculation for $14 p$

Two integers are selected at random from the set ${1, 2, \dots, 11}$ . Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

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Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P (E_{1} ∣ E_{2}) = \frac{1}{2}$ , $P (E_{2} ∣ E_{1}) = \frac{3}{4}$ and $P (E_{1} \cap E_{2}) = \frac{1}{8}$ . Then:

If $\overset{ˉ}{E}$ and $\overset{ˉ}{F}$ are the complementary events of events $E$ and $F$ respectively and if $0 < P (F) < 1$ , then