Class CBSE Class 12 Mathematics Probability Q #909
COMPETENCY BASED
APPLY
3 Marks 2023 SA
The probability distribution of a random variable X is given below :
$$\begin{array}{|c|c|c|c|}
\hline
X & 1 & 2 & 3 \\
\hline
P(X) & \frac{k}{2} & \frac{k}{3} & \frac{k}{6} \\
\hline
\end{array}$$
(i) Find the value of $k$.
(ii) Find $P(1\le X<3)$.
(iii) Find $E(X)$, the mean of $X$.
OR
$A$ and $B$ are independent events such that $P(A\cap\overline{B})=\frac{1}{4}$ and $P(\overline{A}\cap B)=\frac{1}{6}$ Find $P(A)$ and $P(B)$.

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Explanation
The sum of probabilities in a probability distribution must equal $1$.$$\sum P(X) = 1$$$$\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 1$$Find the common denominator (which is $6$):$$\frac{3k}{6} + \frac{2k}{6} + \frac{1k}{6} = 1$$$$\frac{6k}{6} = 1$$$$k = 1$$(ii) Find $P(1 \le X < 3)$The range $1 \le X < 3$ includes $X=1$ and $X=2$, but not $X=3$.$$P(1 \le X < 3) = P(X=1) + P(X=2)$$Substituting $k=1$:$$P(X=1) = \frac{1}{2}, \quad P(X=2) = \frac{1}{3}$$$$P(1 \le X < 3) = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$$(iii) Find $E(X)$, the mean of $X$The expectation (mean) is defined as $E(X) = \sum X \cdot P(X)$.Using $k=1$:$$P(X=1) = \frac{1}{2}, \quad P(X=2) = \frac{1}{3}, \quad P(X=3) = \frac{1}{6}$$$$E(X) = \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{1}{3}\right) + \left(3 \times \frac{1}{6}\right)$$$$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{3}{6}$$$$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{1}{2}$$$$E(X) = 1 + \frac{2}{3} = \frac{5}{3}$$OR Part: Independent EventsProblem:$A$ and $B$ are independent events such that:$P(A \cap \overline{B}) = \frac{1}{4}$$P(\overline{A} \cap B) = \frac{1}{6}$Solution:Let $P(A) = x$ and $P(B) = y$.Since events are independent:$$P(A \cap \overline{B}) = P(A) \cdot P(\overline{B}) = x(1-y)$$$$P(\overline{A} \cap B) = P(\overline{A}) \cdot P(B) = (1-x)y$$We have the system of equations:$x(1-y) = \frac{1}{4} \implies x - xy = \frac{1}{4}$$(1-x)y = \frac{1}{6} \implies y - xy = \frac{1}{6}$Subtracting equation (2) from equation (1):$$(x - xy) - (y - xy) = \frac{1}{4} - \frac{1}{6}$$$$x - y = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}$$$$x = y + \frac{1}{12}$$Substitute $x$ back into equation (2):$$\left(1 - \left(y + \frac{1}{12}\right)\right)y = \frac{1}{6}$$$$\left(\frac{11}{12} - y\right)y = \frac{1}{6}$$$$\frac{11y}{12} - y^2 = \frac{1}{6}$$Multiply by 12 to clear the denominator:$$11y - 12y^2 = 2$$$$12y^2 - 11y + 2 = 0$$Solving the quadratic equation for $y$:$$y = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(12)(2)}}{2(12)}$$$$y = \frac{11 \pm \sqrt{121 - 96}}{24} = \frac{11 \pm \sqrt{25}}{24} = \frac{11 \pm 5}{24}$$Two possible values for $y$:$y_1 = \frac{16}{24} = \frac{2}{3}$$y_2 = \frac{6}{24} = \frac{1}{4}$Finding corresponding $x$ values using $x = y + \frac{1}{12}$:If $y = \frac{2}{3}$: $x = \frac{2}{3} + \frac{1}{12} = \frac{8+1}{12} = \frac{9}{12} = \frac{3}{4}$If $y = \frac{1}{4}$: $x = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}$Answer:There are two possible sets of values:$P(A) = \frac{3}{4}$ and $P(B) = \frac{2}{3}$$P(A) = \frac{1}{3}$ and $P(B) = \frac{1}{4}$

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Step-by-Step Solution

Part (i): Finding the value of k

The sum of probabilities in a probability distribution must equal 1.

Therefore, P(X=1) + P(X=2) + P(X=3) = 1

$$\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 1$$

$$\frac{3k + 2k + k}{6} = 1$$

$$\frac{6k}{6} = 1$$

$$k = 1$$

Part (ii): Finding P(1 ≤ X < 3)

P(1 ≤ X < 3) = P(X=1) + P(X=2)

$$P(1 \le X < 3) = \frac{k}{2} + \frac{k}{3}$$

Since k = 1,

$$P(1 \le X < 3) = \frac{1}{2} + \frac{1}{3} = \frac{3+2}{6} = \frac{5}{6}$$

Part (iii): Finding E(X)

E(X) = Σ [X * P(X)]

E(X) = (1 * P(X=1)) + (2 * P(X=2)) + (3 * P(X=3))

$$E(X) = (1 \times \frac{1}{2}) + (2 \times \frac{1}{3}) + (3 \times \frac{1}{6})$$

$$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{1}{2} = 1 + \frac{2}{3} = \frac{5}{3}$$

OR

Given: $P(A\cap\overline{B})=\frac{1}{4}$ and $P(\overline{A}\cap B)=\frac{1}{6}$

Since A and B are independent events,

$P(A\cap\overline{B}) = P(A)P(\overline{B}) = P(A)(1-P(B)) = \frac{1}{4}$

$P(\overline{A}\cap B) = P(\overline{A})P(B) = (1-P(A))P(B) = \frac{1}{6}$

Let $P(A) = a$ and $P(B) = b$. Then,

$a(1-b) = \frac{1}{4}$ and $(1-a)b = \frac{1}{6}$

$a - ab = \frac{1}{4}$ and $b - ab = \frac{1}{6}$

Subtracting the two equations:

$a - b = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$

$a = b + \frac{1}{12}$

Substituting $a$ in the second equation:

$(1 - (b + \frac{1}{12}))b = \frac{1}{6}$

$(\frac{11}{12} - b)b = \frac{1}{6}$

$\frac{11}{12}b - b^2 = \frac{1}{6}$

$b^2 - \frac{11}{12}b + \frac{1}{6} = 0$

$12b^2 - 11b + 2 = 0$

$(4b - 1)(3b - 2) = 0$

$b = \frac{1}{4}$ or $b = \frac{2}{3}$

If $b = \frac{1}{4}$, $a = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}$

If $b = \frac{2}{3}$, $a = \frac{2}{3} + \frac{1}{12} = \frac{8+1}{12} = \frac{9}{12} = \frac{3}{4}$

Check: If $a = \frac{1}{3}$ and $b = \frac{1}{4}$, $a(1-b) = \frac{1}{3}(1-\frac{1}{4}) = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ and $(1-a)b = (1-\frac{1}{3})\frac{1}{4} = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6}$. This solution is valid.

If $a = \frac{3}{4}$ and $b = \frac{2}{3}$, $a(1-b) = \frac{3}{4}(1-\frac{2}{3}) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$ and $(1-a)b = (1-\frac{3}{4})\frac{2}{3} = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6}$. This solution is also valid.

Correct Answer: k=1, P(1≤X<3) = 5/6, E(X) = 5/3 OR P(A)=1/3, P(B)=1/4 OR P(A)=3/4, P(B)=2/3

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the concepts of probability distribution, expectation, and independent events to solve the problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow specific procedures to calculate the value of k, probabilities, and the expected value. It also involves applying procedures related to independent events.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply probability concepts to solve problems, rather than just recalling definitions or theorems.

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