GSEB Class 12 Statistics Solutions Chapter 2 Random Variable and Discrete Probability Distribution Ex 2.2

Question 1. For a symmetrical binomial distribution with n = 8, find P(X ≤ 1).
Answer: Here, it is given that the binomial distribution is symmetric.
\( \therefore P = q = \frac{1}{2} \), \( n = 8 \)
Putting \( n = 8 \), \( P = \frac{1}{2} \) and \( q = \frac{1}{2} \) into the formula:
\( p(x) = {}^{n}C_x p^x q^{n-x} \)
\( p(x) = {}^{8}C_x \left(\frac{1}{2}\right)^{x}\left(\frac{1}{2}\right)^{8-x} \)
\( P(X \leq 1) = P(X = 0) + P(X = 1) \)
\( = [p(0) + p(1)] \)
\( = [{}^{8}C_0 \left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{8} + {}^{8}C_1 \left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{7}] \)
\( = \left[\left(\frac{1}{2}\right)^{8}+8\left(\frac{1}{2}\right)^{8}\right] \)
\( = \left[\frac{1}{256}+\frac{8}{256}\right] \)
\( = \frac{9}{256} \)
In simple words: When the binomial distribution is balanced, the probability of success and failure are both half. For 8 trials, the chance of having 1 or fewer successes is calculated by adding the probabilities of getting 0 successes and 1 success. This final value turns out to be 9 out of 256.

Exam Tip: Remember that for a symmetrical binomial distribution, the probability of success \( p \) is always equal to 0.5. Clearly show each step of the probability calculation.

Question 2. Mean of a binomial distribution is 5 and its variance is equal to the probability of success. Find the parameters of this distribution and hence find the probability of the event of getting none of the failures for this distribution.
Answer: Here, the mean \( np = 5 \) and variance \( npq = p \) are given.
Putting \( np = 5 \) into the variance equation:
\( npq = p \)
\( 5q = p \)
Putting \( q = 1 - p \):
\( 5(1-p) = p \)
\( 5 - 5p = p \)
\( 5 = p + 5p \)
\( 5 = 6p \)
\( \therefore p = \frac{5}{6} \)
Putting \( p = \frac{5}{6} \) into \( np = 5 \):
\( n \times \frac{5}{6} = 5 \)
\( \therefore n = \frac{5 \times 6}{5} \)
\( \therefore n = 6 \)
So, the parameters of this distribution are \( n = 6 \) and \( p = \frac{5}{6} \).
We can also find \( q \): \( q = 1 - p = 1 - \frac{5}{6} = \frac{1}{6} \)
We need to find the probability of getting none of the failures, which means all successes.
\( \therefore x = 6 \)
Putting \( n = 6 \), \( p = \frac{5}{6} \) and \( q = \frac{1}{6} \) into the probability formula:
\( P(x) = {}^{n}C_x p^x q^{n-x} \)
\( p(x) = {}^{6}C_x \left(\frac{5}{6}\right)^{x} \left(\frac{1}{6}\right)^{6-x} \)
For \( X = 6 \):
\( \therefore P(X = 6) = p(6) = {}^{6}C_6 \left(\frac{5}{6}\right)^{6} \left(\frac{1}{6}\right)^{0} \)
\( = 1 \times \frac{15625}{46656} \times 1 \)
\( = \frac{15625}{46656} \)
Hence, the probability of getting none of the failures for the given distribution is \( \frac{15625}{46656} \).
In simple words: Given the mean and variance, we first find the chances of success \( p \) and the number of trials \( n \). Once these values are known, we can calculate the chance of having zero failures, which means all outcomes were successes. The final probability for all successes is 15625 out of 46656.

Exam Tip: For binomial distributions, remember that mean is \( np \) and variance is \( npq \). Use these formulas to find \( p, q, \) and \( n \) first, then apply the probability mass function \( P(X=x) \) for the specific event.

Question 3. A person has kept 4 cars to run on rent. The probability that any car is rented during the day is 0.6. Find the probability that more than one but less than 4 cars are rented during a day.
Answer: There are 4 cars available for rent, so \( n = 4 \).
Let \( X \) be the number of cars rented. We need to find the probability that more than one but less than 4 cars are rented.
\( \therefore X = 2 \) or \( X = 3 \)
The probability that a car is rented during the day is \( p = 0.6 \).
So, the probability that a car is not rented is \( q = 1 - p = 1 - 0.6 = 0.4 \).
Putting \( n = 4 \), \( p = 0.6 \) and \( q = 0.4 \) into the binomial probability formula:
\( P(X = x) = p(x) = {}^{n}C_x p^x q^{n-x} \)
\( P(X = x) = p(x) = {}^{4}C_x (0.6)^x (0.4)^{4-x} \)
Now, we need to find \( P(X = 2 \text{ or } X = 3) \):
\( = p(2) + p(3) \)
\( = {}^{4}C_2 (0.6)^2 (0.4)^{4-2} + {}^{4}C_3 (0.6)^3 (0.4)^{4-3} \)
\( = 6(0.6)^2 (0.4)^2 + 4(0.6)^3 (0.4)^1 \)
\( = 6(0.36) (0.16) + 4(0.216) (0.4) \)
\( = 0.3456 + 0.3456 \)
\( = 0.6912 \)
Hence, the probability that more than one but less than 4 cars are rented during a day is \( 0.6912 \).
In simple words: A person has 4 cars for rent, and each car has a 0.6 chance of being rented. We want to find the total chance that either 2 or 3 cars get rented. We calculate the individual probabilities for 2 and 3 rented cars using the binomial formula and then add them up. The result shows there's a 0.6912 chance of this happening.

Exam Tip: When calculating probabilities for "more than one but less than \( k \)" or similar ranges, clearly identify the specific integer values of \( x \) that fall within that range. Calculate each individual probability and then sum them up.

Question 4. There are 200 farms in a Taluka. Among the bore wells made in these 200 farms of the Taluka, salted water is found in 20 farms. Find the probability of the event of not getting salted water in 3 out of 5 randomly selected farms from the Taluka.
Answer: Among the bore wells made in 200 farms, salted water is found in 20 farms.
The probability of getting salted water in a farm is \( p = \frac{20}{200} = \frac{1}{10} \).
So, the probability of not getting salted water is \( q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10} \).
Here, we select 5 farms randomly, so \( n = 5 \).
We need to find the probability of not getting salted water in 3 out of 5 farms. This means 3 farms do not have salted water, and therefore 2 farms *do* have salted water.
So, \( X = 2 \) (number of farms with salted water).
Putting \( n = 5 \), \( x = 2 \), \( p = \frac{1}{10} \) and \( q = \frac{9}{10} \) into the binomial probability formula:
\( P(X = x) = p(x) = {}^{n}C_x p^x q^{n-x} \)
\( p(x) = {}^{5}C_x \left(\frac{1}{10}\right)^{x} \left(\frac{9}{10}\right)^{5-x} \)
For \( X = 2 \):
\( \therefore P(X = 2) = p(2) = {}^{5}C_2 \left(\frac{1}{10}\right)^{2} \left(\frac{9}{10}\right)^{5-2} \)
\( = {}^{5}C_2 \left(\frac{1}{10}\right)^{2} \left(\frac{9}{10}\right)^{3} \)
\( = 10 \times \frac{1}{100} \times \frac{729}{1000} \)
\( = 10 \times \frac{1}{100} \times \frac{729}{1000} \)
\( = \frac{7290}{100000} \)
\( = \frac{729}{10000} \)
\( = 0.0729 \)
Hence, the probability of not getting salted water in 3 farms (out of 5) is \( 0.0729 \).
In simple words: Out of 200 farms, 20 have salty water, so the chance of finding salty water is 1 out of 10. We pick 5 farms. We want to find the chance that 3 of these 5 farms *don't* have salty water, which means 2 farms *do* have salty water. Using the binomial formula, this chance comes out to be 0.0729.

Exam Tip: Carefully read the question to define \( p \) (probability of success) and \( x \) (number of successes) correctly. In this case, "not getting salted water in 3 farms" means 3 failures and 2 successes (getting salted water).

Question 5. An example is given to 6 students to solve. The probability of getting correct solution of the problem by any student is 0.6. Students are trying to solve the problem independently. Find the probability of getting the correct solution by only 2 out of the 6 students.
Answer: Here, the number of students is \( n = 6 \).
Let \( x \) be the number of students getting the correct solution. We need only 2 students to get the correct solution.
So, \( x = 2 \).
The probability of getting a correct solution to the problem by any student is \( p = 0.6 \).
Therefore, the probability of not getting a correct solution is \( q = 1 - p = 1 - 0.6 = 0.4 \).
Putting \( n = 6 \), \( x = 2 \), \( p = 0.6 \) and \( q = 0.4 \) into the binomial probability formula:
\( P(X = x) = p(x) = {}^{n}C_x p^x q^{n-x} \)
\( p(x) = {}^{6}C_x (0.6)^x (0.4)^{6-x} \)
For \( X = 2 \):
\( \therefore P(X = 2) = p(2) = {}^{6}C_2 (0.6)^2 (0.4)^{6-2} \)
\( = {}^{6}C_2 (0.6)^2 (0.4)^4 \)
\( = 15(0.36) (0.0256) \)
\( = 0.13824 \)
Hence, the probability of getting the correct solution from only 2 out of the 6 students is \( 0.13824 \).
In simple words: There are 6 students, and each has a 0.6 chance of solving a problem correctly. We want to find the chance that exactly 2 out of these 6 students get the right answer. We use the binomial probability formula with 6 trials, 2 successes, and the given success/failure rates. The probability for this exact outcome is 0.13824.

Exam Tip: Clearly identify \( n \) (total trials), \( p \) (probability of success in one trial), and \( x \) (desired number of successes) before applying the binomial probability formula. Pay attention to the exponent values for \( p \) and \( q \).