JEE Advanced 2011

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $11/25$ and the probability of none of them occurring is $2/25$ . If $P (T)$ denotes the probability of occurrence of the event $T$ , then

Q: Why are there two possible answers for $P(E)$ and $P(F)$?

Because the problem is symmetric with respect to $E$ and $F$. The equations we derived, $x+y = \frac{7}{5}$ and $xy = \frac{12}{25}$, do not distinguish between $x$ and $y$. Therefore, if $(x, y) = (4/5, 3/5)$ is a solution, then $(x, y) = (3/5, 4/5)$ must also be a solution. Both are perfectly valid!

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: Visualize the sample space as a Venn diagram with two overlapping circles, E and F. The region representing 'exactly one' is the sum of the parts of E and F that do not overlap, which can be expressed as

P (E) + P (F) - 2 P (E \cap F)

. The region representing 'none' is the area outside both circles, which is

1 - P (E \cup F)

Hint 2

Hint 2: Leverage the power of independence. For independent events, the probability of their intersection is the product of their individual probabilities:

P (E \cap F) = P (E) P (F)

. Let

P (E) = x

and

P (F) = y

. Now, rewrite your expressions from Hint 1 in terms of

x

and

y

Hint 3

Hint 3: You now have a system of two equations:

x + y - 2 x y = \frac{11}{25}

and

(1 - x) (1 - y) = \frac{2}{25}

. Expand the second equation to get

x + y - x y = \frac{23}{25}

. Subtract the first equation from this new one to isolate

x y

, then find

x + y

and solve the quadratic equation

t^{2} - (x + y) t + x y = 0

Visualized Solution

Defining the Probabilities

Let the probability of event $E$ be $P (E) = x$ .
Let the probability of event $F$ be $P (F) = y$ .
Our goal is to find the values of $x$ and $y$ using the given information.

The Concept of Independence

Since $E$ and $F$ are independent events:
The probability of both occurring is the product of their individual probabilities.
$P (E \cap F) = P (E) \cdot P (F) = x \cdot y$

Equation for Exactly One Event

Probability that exactly one occurs is $\frac{11}{25}$ .
This means either $E$ occurs without $F$ , or $F$ occurs without $E$ .
$P (E) + P (F) - 2 P (E \cap F) = \frac{11}{25}$
$x + y - 2 x y = \frac{11}{25} \dots (1)$

Equation for None of the Events

Probability that none occurs is $\frac{2}{25}$ .
This is the region outside both circles.
$P (E^{c} \cap F^{c}) = (1 - x) (1 - y) = \frac{2}{25}$
$1 - x - y + x y = \frac{2}{25}$
$x + y - x y = 1 - \frac{2}{25} = \frac{23}{25} \dots (2)$

Solving for the Product $x y$

We have two equations:
1) $x + y - 2 x y = \frac{11}{25}$
2) $x + y - x y = \frac{23}{25}$
Subtract equation (1) from equation (2):
$(x + y - x y) - (x + y - 2 x y) = \frac{23}{25} - \frac{11}{25}$
$x y = \frac{12}{25}$

Solving for the Sum $x + y$

Substitute $x y = \frac{12}{25}$ into equation (2):
$x + y - \frac{12}{25} = \frac{23}{25}$
$x + y = \frac{23}{25} + \frac{12}{25}$
$x + y = \frac{35}{25} = \frac{7}{5}$

Forming the Quadratic Equation

We know the sum ( $x + y = \frac{7}{5}$ ) and product ( $x y = \frac{12}{25}$ ).
$x$ and $y$ are the roots of the quadratic equation:
$t^{2} - (Sum) t + (Product) = 0$
$t^{2} - \frac{7}{5} t + \frac{12}{25} = 0$
Multiply by $25$ : $25 t^{2} - 35 t + 12 = 0$

Factoring and Finding Roots

Factorize the quadratic equation:
$25 t^{2} - 20 t - 15 t + 12 = 0$
$5 t (5 t - 4) - 3 (5 t - 4) = 0$
$(5 t - 4) (5 t - 3) = 0$
$t = \frac{4}{5}$ or $t = \frac{3}{5}$

Final Answer and Options

The set of probabilities is ${P (E), P (F)} = {\frac{4}{5}, \frac{3}{5}}$ .
Case 1: $P (E) = \frac{4}{5}, P (F) = \frac{3}{5}$ (Matches Option A)
Case 2: $P (E) = \frac{3}{5}, P (F) = \frac{4}{5}$ (Matches Option D)
Therefore, both (A) and (D) are correct.

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The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Setup

We are exploring the probabilities of two events,

E

and

F

, denoted as

x

and

y

respectively. We are given that these events are independent, which provides a powerful tool for our calculations.

For independent events, the probability of their intersection is the product of their individual probabilities:

P (E \cap F) = P (E) \cdot P (F) = x y

Translating the Clues into Algebra

The probability that exactly one event occurs is given as

\frac{11}{25}

. This corresponds to the sum of the probabilities of

E

occurring without

F

and

F

occurring without

E

x + y - 2 x y = \frac{11}{25}

The probability that neither event occurs is given as

\frac{2}{25}

. Since the events are independent, their complements are also independent, leading to:

P (E^{c} \cap F^{c}) = (1 - x) (1 - y) = \frac{2}{25}

Expanding this expression, we obtain:

1 - x - y + x y = \frac{2}{25} \Rightarrow x + y - x y = \frac{23}{25}

The Algebraic Dance

We now have a system of two equations: 1)

x + y - 2 x y = \frac{11}{25}

x + y - x y = \frac{23}{25}

Subtracting the first equation from the second eliminates the sum

(x + y)

(x + y - x y) - (x + y - 2 x y) = \frac{23}{25} - \frac{11}{25}

x y = \frac{12}{25}

Substituting

x y = \frac{12}{25}

back into the second equation allows us to solve for the sum:

x + y = \frac{23}{25} + \frac{12}{25} = \frac{35}{25} = \frac{7}{5}

The Quadratic Bridge

Given the sum and product of

x

and

y

, we can construct a quadratic equation

t^{2} - (sum) t + (product) = 0

to find the individual probabilities:

t^{2} - \frac{7}{5} t + \frac{12}{25} = 0

Multiplying by

25

to clear the denominators, we get:

25 t^{2} - 35 t + 12 = 0

Factoring the quadratic equation:

(5 t - 4) (5 t - 3) = 0

This yields the roots

t = \frac{4}{5}

and

t = \frac{3}{5}

. Therefore, the probabilities of the two events are $\frac{4}{5}$ and $\frac{3}{5}$ .

Common Traps & FAQs

Wait, why do we subtract

2 P (E \cap F)

when calculating the probability of exactly one event?

Great question! Think of the Venn diagram. The region 'E only' is

P (E) - P (E \cap F)

, and 'F only' is

P (F) - P (E \cap F)

. When you add these two regions together to get 'exactly one', you get

P (E) + P (F) - 2 P (E \cap F)

. We subtract the intersection twice because it is included once in

P (E)

and once in

P (F)

, but we want to remove it entirely from our calculation.

Why can't we just use energy conservation or some other shortcut here?

Probability problems are essentially puzzles of logical structure. While there isn't a 'conservation law' like in physics, the 'conservation of probability' (the sum of all disjoint probabilities equals 1) is exactly what we used to define the 'none' region! The shortcut here is recognizing the symmetry of

x

and

y

in the equations, which allows us to treat them as roots of a quadratic equation.

Why are there two possible answers for

P (E)

and

P (F)

Because the problem is symmetric with respect to

E

and

F

. The equations we derived,

x + y = \frac{7}{5}

and

x y = \frac{12}{25}

, do not distinguish between

x

and

y

. Therefore, if

(x, y) = (4/5, 3/5)

is a solution, then

(x, y) = (3/5, 4/5)

must also be a solution. Both are perfectly valid!

Animated Solution for Mathematics - Probability: Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $11/25$ and the probability of none of them occurring is $2/25$ . If $P (T)$ denotes the probability of occurrence of the event $T$ , then

Select Answer:

Visualized Solution

Defining the Probabilities

The Concept of Independence

Equation for Exactly One Event

Equation for None of the Events

Solving for the Product $x y$

Solving for the Sum $x + y$

Forming the Quadratic Equation

Factoring and Finding Roots

Final Answer and Options

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

Translating the Clues into Algebra

The Algebraic Dance

The Quadratic Bridge

Similar Questions

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Select Answer:

Visualized Solution

Defining the Probabilities

The Concept of Independence

Equation for Exactly One Event

Equation for None of the Events

Solving for the Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xy

Solving for the Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+y

Forming the Quadratic Equation

Factoring and Finding Roots

Final Answer and Options

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

Translating the Clues into Algebra

The Algebraic Dance

The Quadratic Bridge

Similar Questions

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