JEE Main 2021 (27 Aug Shift 2)

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Animated Solution for Mathematics - Probability: Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :

Q: Why are we squaring the probabilities in the final step?

Because the events are independent. If A gets $r$ heads and B gets $r$ heads, the joint probability is the product of the individual probabilities. Since the distributions are identical, $P(X=r) \cdot P(Y=r)$ becomes $[P(X=r)]^2$. We are essentially summing the probabilities of all mutually exclusive scenarios where they match.

Select Answer:

Progressive Hints

Hint 1

Visualize the experiment as two independent branches of a probability tree. You are not just looking for one outcome, but the intersection of two identical probability distributions. Do not rush to calculate the total sample space of 64; instead, focus on the distribution of heads for a single person first.

Hint 2

Recall the Binomial Distribution formula:

P (X = r) = (r n) p^{r} q^{n - r}

. Since the coins are fair,

p = q = 1/2

. Calculate the probability distribution for a single person for

r = 0, 1, 2, 3

. This will give you the weights for each possible outcome.

Hint 3

The condition

P (X = Y)

is satisfied when both get 0, 1, 2, or 3 heads. Since the events are independent,

P (X = r \cap Y = r) = P (X = r) \cdot P (Y = r)

. Your starting equation is

P (X = Y) = \sum_{r = 0}^{3} [P (X = r)]^{2}

. Now, substitute your calculated values into this summation.

Visualized Solution

Visualizing the Experiment

Two independent experimenters: A and B.
Each tosses $3$ fair coins.
Total outcomes for each person = $2^{3} = 8$ .

Defining Random Variables $X$ and $Y$

Let $X$ = Number of heads obtained by A.
Let $Y$ = Number of heads obtained by B.
Possible values: $X, Y \in {0, 1, 2, 3}$ .

The Binomial Probability Tool

Probability of heads $p = \frac{1}{2}$ , tails $q = \frac{1}{2}$ .
$P (X = r) = (r 3) p^{r} q^{3 - r} = (r 3) (\frac{1}{2})^{3}$
$P (X = r) = \frac{( r 3 )}{8}$

Calculating Individual Probabilities

$P (X = 0) = \frac{( 0 3 )}{8} = \frac{1}{8}$
$P (X = 1) = \frac{( 1 3 )}{8} = \frac{3}{8}$
$P (X = 2) = \frac{( 2 3 )}{8} = \frac{3}{8}$
$P (X = 3) = \frac{( 3 3 )}{8} = \frac{1}{8}$

The Condition for Equality

We need to find $P (X = Y)$ .
This occurs when $(X = 0, Y = 0) \cup (X = 1, Y = 1) \cup (X = 2, Y = 2) \cup (X = 3, Y = 3)$ .
Since events are independent: $P (X = r, Y = r) = P (X = r) \cdot P (Y = r)$ .

Setting up the Summation

$P (X = Y) = \sum_{r = 0}^{3} P (X = r) \cdot P (Y = r)$
Since $P (X = r) = P (Y = r)$ , then:
$P (X = Y) = \sum_{r = 0}^{3} [P (X = r)]^{2}$

Substitution of Values

$P (X = Y) = (\frac{1}{8})^{2} + (\frac{3}{8})^{2} + (\frac{3}{8})^{2} + (\frac{1}{8})^{2}$

Squaring the Terms

$P (X = Y) = \frac{1}{64} + \frac{9}{64} + \frac{9}{64} + \frac{1}{64}$

Summing the Fractions

$P (X = Y) = \frac{1 + 9 + 9 + 1}{64}$
$P (X = Y) = \frac{20}{64}$

Simplifying the Result

Divide numerator and denominator by $4$ :
$\frac{20 \div 4}{64 \div 4} = \frac{5}{16}$
Final Probability = $\frac{5}{16}$

Key Takeaway and Summary

Key Takeaway: For independent identical experiments, $P (X = Y) = \sum [P (X = r)]^{2}$ .
Final Answer: Option (3) which is $\frac{5}{16}$ .

00:00 / 00:00

The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Defining the Variables

Let

X

be the number of heads obtained by friend

A

, and

Y

be the number of heads obtained by friend

B

. Since each person tosses three fair coins, the possible values for both

X

and

Y

are

{0, 1, 2, 3}

We utilize the Binomial Probability formula to determine the likelihood of each outcome:

P (X = r) = (r 3) (\frac{1}{2})^{3} = \frac{( r 3 )}{8}

Calculating Individual Probabilities

Given that the probability of heads

p

1/2

and tails

q

1/2

, we calculate the probability for each possible value of

r

For

r = 0

P (X = 0) = \frac{( 0 3 )}{8} = \frac{1}{8}

For

r = 1

P (X = 1) = \frac{( 1 3 )}{8} = \frac{3}{8}

For

r = 2

P (X = 2) = \frac{( 2 3 )}{8} = \frac{3}{8}

For

r = 3

P (X = 3) = \frac{( 3 3 )}{8} = \frac{1}{8}

The Master Equation for $P (X = Y)$

The event

X = Y

occurs if both friends obtain the same number of heads. Because

A

and

B

are independent, the probability of a joint event is the product of their individual probabilities.

We express the total probability as:

P (X = Y) = P (X = 0) P (Y = 0) + P (X = 1) P (Y = 1) + P (X = 2) P (Y = 2) + P (X = 3) P (Y = 3)

Since the distributions are identical, this simplifies to the sum of the squares:

P (X = Y) = r = 0 \sum 3 [P (X = r)]^{2}

Final Calculation

Substituting our calculated values into the summation:

P (X = Y) = (\frac{1}{8})^{2} + (\frac{3}{8})^{2} + (\frac{3}{8})^{2} + (\frac{1}{8})^{2}

P (X = Y) = \frac{1}{64} + \frac{9}{64} + \frac{9}{64} + \frac{1}{64} = \frac{20}{64}

By simplifying the fraction

\frac{20}{64}

by a factor of 4, we arrive at the final result:

$P (X = Y) = \frac{5}{16}$

Common Traps & FAQs

Why can't I just say the probability is

1/4

because there are 4 possible outcomes (0, 1, 2, 3 heads)?

That is a classic trap! You are assuming that getting 0 heads is just as likely as getting 1 head. In reality, the outcomes 0, 1, 2, and 3 heads are not equally likely. You must weight them by their binomial probabilities. Always check if your sample space outcomes are equiprobable before using simple counting methods.

Why are we squaring the probabilities in the final step?

Because the events are independent. If A gets

r

heads and B gets

r

heads, the joint probability is the product of the individual probabilities. Since the distributions are identical,

P (X = r) \cdot P (Y = r)

becomes

[P (X = r)]^{2}

. We are essentially summing the probabilities of all mutually exclusive scenarios where they match.

Is there a faster way to solve this without calculating all the cases?

While you could use symmetry arguments, calculating the individual probabilities is the safest and most robust method for JEE Advanced. It prevents silly errors. The symmetry here is beautiful—notice how

P (0) = P (3)

and

P (1) = P (2)

—but writing out the sum ensures you don't miss any terms.

Animated Solution for Mathematics - Probability: Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :

Select Answer:

Visualized Solution

Visualizing the Experiment

Defining Random Variables $X$ and $Y$

The Binomial Probability Tool

Calculating Individual Probabilities

The Condition for Equality

Setting up the Summation

Substitution of Values

Squaring the Terms

Summing the Fractions

Simplifying the Result

Key Takeaway and Summary

The Sigma Insight: Random Variables and Probability Distributions

Defining the Variables

Calculating Individual Probabilities

The Master Equation for $P (X = Y)$

Final Calculation

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Animated Solution for Mathematics - Probability: Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :

Select Answer:

Visualized Solution

Visualizing the Experiment

Defining Random Variables .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

The Binomial Probability Tool

Calculating Individual Probabilities

The Condition for Equality

Setting up the Summation

Substitution of Values

Squaring the Terms

Summing the Fractions

Simplifying the Result

Key Takeaway and Summary

The Sigma Insight: Random Variables and Probability Distributions

Defining the Variables

Calculating Individual Probabilities

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

Final Calculation

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A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-

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