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 $P(B)$.

Prev Next
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 $P(B) = \frac{1}{4}$

AI Tutor Explanation

Powered by Gemini

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

AI generated content. Review strictly for academic accuracy.

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.

More from this Chapter

SA
31. From a lot of 30 bulbs which include 6 defective bulbs, a sample of 2 bulbs is drawn at random one by one with replacement. Find the probability distribution of the number of defective bulbs and hence find the mean number of defective bulbs.
SA
The chances of P, Q and R getting selected as CEO of a company are in the ratio 4: 1: 2 respectively. The probabilities for the company to increase its profits from the previous year under the new CEO, P, Q or R are 0-3, 0-8 and 0.5 respectively. If the company increased the profits from the previous year, find the probability that it is due to the appointment of R as CEO.
SA
A biased die is twice as likely to show an even number as an odd number. If such a die is thrown twice, find the probability distribution of the number of sixes. Also, find the mean of the distribution.
SA
A person is Head of two independent selection committees I and II. If the probability of making a wrong selection in committee I is 0.03 and that in committee II is 0.01, then find the probability that the person makes the correct decision of selection: (i) in both committees (ii) in only one committee.
SA
For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in written test is 0.4 and that a female getting a distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test.
View All Questions