JEE Advanced 1992

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Animated Solution for Mathematics - Probability: 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

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Visualized Solution

Understanding the Match Scenario

Total matches played: $n = 4$ (2 with West Indies, 2 with Australia)
Points per match $X \in {0, 1, 2}$
Probabilities: $P (X = 0) = 0.45$ , $P (X = 1) = 0.05$ , $P (X = 2) = 0.50$

Defining the Target Event

Maximum possible points: $4 \times 2 = 8$ points
We need "at least 7 points"
This means the total points $S$ must be either $7$ or $8$
Required Probability: $P (S \geq 7) = P (S = 7) + P (S = 8)$

Case 1: Getting Exactly 8 Points

To get $S = 8$ , India must win maximum points in all matches
Points in 4 matches must be $(2, 2, 2, 2)$
Probability of getting 2 points in one match: $P (2) = 0.50$

Calculating $P (S = 8)$

Since matches are independent, we multiply the probabilities:
$P (S = 8) = P (2) \times P (2) \times P (2) \times P (2)$
$P (S = 8) = (0.50)^{4}$
$P (S = 8) = 0.0625$

Case 2: Getting Exactly 7 Points

To get $S = 7$ , we need a combination of points summing to 7
The only combination using ${0, 1, 2}$ is three $2$ s and one $1$
Possible set of points: ${2, 2, 2, 1}$

Arrangements for $S = 7$

The single 1-point match can occur in any of the 4 matches
Number of arrangements $= \frac{4 !}{3 ! 1 !} = 4$
The 4 sequences are: $(1, 2, 2, 2)$ , $(2, 1, 2, 2)$ , $(2, 2, 1, 2)$ , $(2, 2, 2, 1)$

Calculating $P (S = 7)$

Probability of one such sequence: $(0.50)^{3} \times (0.05)^{1}$
$P (S = 7) = 4 \times [P (2)]^{3} \times [P (1)]^{1}$
$P (S = 7) = 4 \times (0.50)^{3} \times (0.05) = 0.0250$

Final Probability Calculation

Required Probability $P (S \geq 7) = P (S = 8) + P (S = 7)$
$P (S \geq 7) = 0.0625 + 0.0250$
Final Result $= 0.0875$

Key Takeaways and Conclusion

Key Takeaway 1: Break 'at least' conditions into mutually exclusive cases.
Key Takeaway 2: Use permutations (like $\frac{n !}{r _{1} ! r _{2} ! ...}$ ) when multiple matches have identical point outcomes.
Next Challenge: Try calculating the probability for at least 6 points. How many new cases appear?

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The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Setup

We are given

n = 4

independent matches. The points scored in each match,

X_{i}

, follow the distribution:

P (X = 0) = 0.45

P (X = 1) = 0.05

, and

P (X = 2) = 0.50

The total points

S

is defined as the sum of points from each match:

S = X_{1} + X_{2} + X_{3} + X_{4}

Since the maximum points per match is

2

, the maximum possible total is

4 \times 2 = 8

. We are tasked with finding the probability that India secures at least

7

points, denoted as

P (S \geq 7)

The Master Equation

Because the maximum possible score is

8

, the condition

S \geq 7

is satisfied only by two mutually exclusive cases:

S = 7

S = 8

Therefore, the total probability is:

P (S \geq 7) = P (S = 8) + P (S = 7)

Case 1

The Perfect Streak (

S = 8

)

To achieve a total of

8

points, every match must result in

2

points. The only possible sequence is

(2, 2, 2, 2)

Since the matches are independent, we calculate the probability as:

P (S = 8) = P (2) \times P (2) \times P (2) \times P (2) = (0.50)^{4}

P (S = 8) = 0.0625

Case 2

The Near-Perfect Streak (

S = 7

)

To achieve a total of

7

points, the set of scores must be

{2, 2, 2, 1}

. The order of these scores matters, as the single match resulting in

1

point can occur in any of the

4

positions.

The number of distinct arrangements is given by the permutation formula:

\frac{4 !}{3 ! 1 !} = 4

Each of these

4

sequences has a probability of:

P (2)^{3} \times P (1)^{1} = (0.50)^{3} \times (0.05) = 0.125 \times 0.05 = 0.00625

Thus, the total probability for

S = 7

is:

P (S = 7) = 4 \times 0.00625 = 0.0250

Final Calculation

We now combine the probabilities of these two mutually exclusive events to find the final result:

P (S \geq 7) = 0.0625 + 0.0250

P (S \geq 7) = 0.0875

The probability that India secures at least

7

points is $0.0875$ .

Common Traps & FAQs

Wait, why can't we just use the binomial distribution formula here?

That is a fantastic question! The binomial distribution is only applicable when there are exactly two outcomes per trial (success or failure). Here, we have three possible outcomes (

0, 1, 2

points) with different probabilities. Because the outcomes are not binary, we must treat this as a multinomial scenario or simply break it down into specific, mutually exclusive cases.

Why do we multiply the probability of the

S = 7

case by 4?

Think of the four matches as distinct slots: Match 1, Match 2, Match 3, and Match 4. To get a total of

7

points, we need three matches to result in

2

points and one match to result in

1

point. The '1-point' match could be any of the four. Since there are

4

distinct positions for that single '1', there are

4

different sequences (e.g.,

1, 2, 2, 2

2, 1, 2, 2

, etc.). Each sequence has the same probability, so we multiply the probability of one sequence by

4

Is there a faster way to solve this than listing cases?

Yes, for more complex problems, you could use the concept of 'Generating Functions'. You would define a polynomial for one match:

P (x) = 0.45 x^{0} + 0.05 x^{1} + 0.50 x^{2}

. Then, the probability distribution for four matches is given by the coefficients of

[P (x)]^{4}

. However, for a small number of matches like

4

, listing the cases is often faster and less prone to algebraic errors!

Select Answer:

Visualized Solution

Understanding the Match Scenario

Defining the Target Event

Case 1: Getting Exactly 8 Points

Calculating $P (S = 8)$

Case 2: Getting Exactly 7 Points

Arrangements for $S = 7$

Calculating $P (S = 7)$

Final Probability Calculation

Key Takeaways and Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Master Equation

Case 1

Case 2

Final Calculation

Similar Questions

Comprehension Passage

$P (X > Y)$ is

$P (X = Y)$ is

List-I

List-II

The probability that at least one of the events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.2$ , then $P (\overset{ˉ}{A}) + P (\overset{ˉ}{B})$ is

Four persons can hit a target correctly with probabilities $1/2, 1/3, 1/4$ and $1/8$ respectively. if all hit at the target independently, then the probability that the target would be hit, is

For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = $\frac{1}{4}$ and P(All the three events occur simultaneously) = $\frac{1}{16}$ . Then the probability that at least one of the events occurs, is:

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is

Select Answer:

Visualized Solution

Understanding the Match Scenario

Defining the Target Event

Case 1: Getting Exactly 8 Points

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(S=8)

Case 2: Getting Exactly 7 Points

Arrangements for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S=7

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(S=7)

Final Probability Calculation

Key Takeaways and Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Master Equation

Case 1

Case 2

Final Calculation

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

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then P(Aˉ)+P(Bˉ) is

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is

Calculating $P (S = 8)$

Arrangements for $S = 7$

Calculating $P (S = 7)$

$P (X > Y)$ is

$P (X = Y)$ is

The probability that at least one of the events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.2$ , then $P (\overset{ˉ}{A}) + P (\overset{ˉ}{B})$ is