RBSE Solutions Class 12 Maths Chapter 16 Probability and Probability Distribution Misc.

 

Question 1. Two events A and B are said to be mutually independent, if:
(i) \( P(A) = P(B) \)
(ii) \( P(A) + P(B) = 1 \)
(iii) \( P(\overline{A} \overline{B}) = [1-P(A)] [1-P(B)] \)
(iv) A and B are mutually disjoint
Answer: (iii) \( P(\overline{A} \overline{B}) = [1-P(A)] [1-P(B)] \)
In simple words: For two independent events, the probability that neither A nor B happens is the product of their individual probabilities of not happening. This means \( P(\overline{A} \cap \overline{B}) = P(\overline{A}) \times P(\overline{B}) \).

🎯 Exam Tip: Remember that for independent events, \( P(A \cap B) = P(A)P(B) \), and similarly for their complements, \( P(\overline{A} \cap \overline{B}) = P(\overline{A})P(\overline{B}) \).

 

Question 2. Two dice are thrown. The probability of getting a pair of even prime number is :
(i) \( \frac {1}{3} \)
(ii) 0
(iii) \( \frac {1}{36} \)
(iv) \( \frac {1}{12} \)
Answer: (iii) \( \frac {1}{36} \)
In simple words: The only even prime number is 2. So, we want both dice to show a 2. The chance of rolling a 2 on one die is 1 out of 6. Since the rolls are independent, the chance of rolling two 2s is \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \).

🎯 Exam Tip: Always identify the sample space and the specific event outcomes clearly. The number 2 is unique as the only even prime number, which simplifies calculations here.

 

Question 3. If A and B be two events, such that A \( \subset \) B and \( P(B) \neq 0 \), then which of the following statement is true?
(i) \( P(\frac{A}{B}) < P(A) \)
(ii) \( P(\frac{A}{B}) \ge P(A) \)
(iii) \( P(\frac{A}{B}) = P(A) \)
(iv) None of these
Answer: (ii) \( P(\frac{A}{B}) \ge P(A) \)
In simple words: If event A is a subset of event B (A happens only if B also happens), then the probability of A happening given B has happened must be greater than or equal to the probability of A happening alone. This is because knowing B has occurred increases the chances for A if A is contained within B.

🎯 Exam Tip: When \( A \subset B \), then \( P(A \cap B) = P(A) \). The conditional probability \( P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} \). Since \( P(B) \le 1 \), it implies \( \frac{1}{P(B)} \ge 1 \), so \( P(A|B) \ge P(A) \).

 

Question 4. Two cards are drawn randomly from a well- shuffled deck of 52 cards. If X denotes the number of aces, then find mean of X:
(i) \( \frac{5}{13} \)
(ii) \( \frac{1}{13} \)
(iii) \( \frac{37}{221} \)
(iv) \( \frac{2}{13} \)
Answer: (iv) \( \frac{2}{13} \)
Solution:
Total number of cards = 52
Total number of aces = 4
Total number of non-aces = 48
Number of ways to draw 2 cards from 52 is \( ^{52}C_2 = \frac{52 \times 51}{2} = 1326 \)
Let X be the number of aces drawn. X can take values 0, 1, or 2.
For \( X=0 \): Both cards are non-aces.
Number of ways to draw 2 non-aces from 48 is \( ^{48}C_2 = \frac{48 \times 47}{2} = 1128 \)
\( P(X=0) = \frac{1128}{1326} \)
For \( X=1 \): One ace and one non-ace.
Number of ways to draw 1 ace from 4 is \( ^{4}C_1 = 4 \)
Number of ways to draw 1 non-ace from 48 is \( ^{48}C_1 = 48 \)
Number of ways to draw 1 ace and 1 non-ace is \( 4 \times 48 = 192 \)
\( P(X=1) = \frac{192}{1326} \)
For \( X=2 \): Both cards are aces.
Number of ways to draw 2 aces from 4 is \( ^{4}C_2 = \frac{4 \times 3}{2} = 6 \)
\( P(X=2) = \frac{6}{1326} \)
The probability distribution is:

Now, the mean \( E(X) = \sum X \cdot P(X) \)
\( E(X) = (0 \times \frac{1128}{1326}) + (1 \times \frac{192}{1326}) + (2 \times \frac{6}{1326}) \)
\( E(X) = 0 + \frac{192}{1326} + \frac{12}{1326} \)
\( E(X) = \frac{192 + 12}{1326} = \frac{204}{1326} \)
We can simplify the fraction:
\( \frac{204}{1326} = \frac{102}{663} = \frac{34}{221} = \frac{2}{13} \)
So, \( E(X) = \frac{2}{13} \).
In simple words: We calculate the chance of getting zero aces, one ace, or two aces when drawing two cards. Then, we multiply each possible number of aces by its probability and add these values together. This sum gives us the average number of aces we expect to draw, which is 2/13.

🎯 Exam Tip: For calculating the mean of a random variable, remember the formula \( E(X) = \sum x_i P(X=x_i) \). Always ensure your probabilities sum to 1 before calculating the mean.

 

Question 5. A random variable X, takes the values 0,1,2, and 3. Mean of X is \( P(x = 3) = 2P(x = 1) \) and \( P(x = 2) = 0.3 \), then \( P(X=0) \) is :
(i) 0.2
(ii) 0.4
(iii) 0.3
(iv) 0.1
Answer: (ii) 0.4
Solution:
Let \( P(X=0) = x \)
Let \( P(X=1) = p \)
We are given \( P(X=2) = 0.3 \)
We are given \( P(X=3) = 2P(X=1) = 2p \)
The mean of X is given as 1.3. (This value is derived from the source's calculation on page 5, which states `1.3 = x*0 + 1*p + 2*0.3 + 3*2p`).
So, the probability distribution is:

The sum of all probabilities must be 1:
\( P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1 \)
\( x + p + 0.3 + 2p = 1 \)
\( x + 3p + 0.3 = 1 \)
\( x + 3p = 1 - 0.3 \)
\( x + 3p = 0.7 \) ... (1)

Now, calculate the mean \( E(X) = \sum X \cdot P(X) \). From the source, the mean is 1.3.
\( E(X) = (0 \times x) + (1 \times p) + (2 \times 0.3) + (3 \times 2p) \)
\( 1.3 = 0 + p + 0.6 + 6p \)
\( 1.3 = 7p + 0.6 \)
\( 7p = 1.3 - 0.6 \)
\( 7p = 0.7 \)
\( p = \frac{0.7}{7} \)
\( p = 0.1 \)

Substitute the value of \( p \) into equation (1):
\( x + 3(0.1) = 0.7 \)
\( x + 0.3 = 0.7 \)
\( x = 0.7 - 0.3 \)
\( x = 0.4 \)
Thus, \( P(X=0) = 0.4 \).
In simple words: We are given parts of a probability distribution and its mean. We use the fact that all probabilities must add up to 1, and the formula for the mean of a discrete random variable, to create two equations. By solving these equations, we find the unknown probability \( P(X=0) \).

🎯 Exam Tip: Always set up a system of equations: one for the sum of probabilities being 1, and another for the given mean. This method is crucial for finding unknown probabilities in a distribution.

 

Question 6. Probability of girl to be a racer is \( \frac {4}{5} \). What is the probability that 4 girls out of 5 are racer ?
Answer:
This is a binomial probability problem.
Let \( n \) be the total number of girls selected, \( n=5 \).
Let \( p \) be the probability that a girl is a racer, \( p = \frac{4}{5} \).
Let \( q \) be the probability that a girl is not a racer, \( q = 1 - p = 1 - \frac{4}{5} = \frac{1}{5} \).
We want to find the probability that exactly 4 girls out of 5 are racers, i.e., \( P(X=4) \).
Using the binomial probability formula \( P(X=k) = \binom{n}{k} p^k q^{n-k} \):
\( P(X=4) = \binom{5}{4} \left(\frac{4}{5}\right)^4 \left(\frac{1}{5}\right)^{5-4} \)
\( P(X=4) = \binom{5}{4} \left(\frac{4}{5}\right)^4 \left(\frac{1}{5}\right)^1 \)
\( P(X=4) = 5 \times \frac{256}{625} \times \frac{1}{5} \)
\( P(X=4) = \frac{256}{625} \)
In simple words: We are looking for the chance that 4 out of 5 girls are racers when we know the individual chance of being a racer. We use a special formula that combines the number of ways this can happen with the probabilities of success and failure for each girl.

🎯 Exam Tip: Recognize binomial distribution problems when there's a fixed number of trials, two possible outcomes (success/failure), and a constant probability of success for each trial. The formula \( \binom{n}{k} p^k q^{n-k} \) is key.

 

Question 7. In a box containing 100, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is :
(i) \( \left(\frac{1}{2}\right)^5 \)
(ii) \( 10^{-1} \)
(iii) \( \left(\frac{9}{10}\right) \)
(iv) \( \left(\frac{9}{10}\right)^5 \)
Answer: (iv) \( \left(\frac{9}{10}\right)^5 \)
Solution:
Total number of bulbs = 100
Number of defective bulbs = 10
Probability of a bulb being defective \( P(D) = \frac{10}{100} = \frac{1}{10} \)
Probability of a bulb being non-defective \( P(\overline{D}) = 1 - P(D) = 1 - \frac{1}{10} = \frac{9}{10} \)
We are taking a sample of 5 bulbs and want the probability that none of them are defective. This means all 5 bulbs are non-defective.
Assuming the selections are independent (or sampling with replacement, or sample size is small compared to total population for without replacement):
\( P(\text{none defective}) = P(\overline{D}) \times P(\overline{D}) \times P(\overline{D}) \times P(\overline{D}) \times P(\overline{D}) \)
\( P(\text{none defective}) = \left(\frac{9}{10}\right)^5 \)
In simple words: We find the chance that one bulb is not broken. Since we want five bulbs not to be broken, and each bulb's condition is separate, we multiply this chance by itself five times.

🎯 Exam Tip: Clearly distinguish between defective and non-defective probabilities. For "none defective" in a sample, it means all items must be non-defective, so multiply the non-defective probability for each item.

 

Question 8. A couple has two children. Find the probability that
(i) both children are male, if it is known that the elder child is a male.
(ii) both children are female, if it is known that the elder child is a female.
(iii) both children are male, if it is known that at least one of the children is male.
Answer:
Let's list the possible outcomes for two children, with M for male and F for female. Assuming equal probability for each sex, the sample space is: {MM, MF, FM, FF}, each with probability \( \frac{1}{4} \).

(i) Both children are male, if it is known that the elder child is a male.
Let A be the event "both children are male" = {MM}.
Let B be the event "elder child is a male" = {MM, MF}.
We need to find \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).
The intersection \( A \cap B \) is {MM}. So, \( P(A \cap B) = \frac{1}{4} \).
The probability \( P(B) = P(\{MM, MF\}) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \).
Therefore, \( P(A|B) = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \).

(ii) Both children are female, if it is known that the elder child is a female.
Let A be the event "both children are female" = {FF}.
Let B be the event "elder child is a female" = {FM, FF}.
We need to find \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).
The intersection \( A \cap B \) is {FF}. So, \( P(A \cap B) = \frac{1}{4} \).
The probability \( P(B) = P(\{FM, FF\}) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \).
Therefore, \( P(A|B) = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \).

(iii) Both children are male, if it is known that at least one of the children is male.
Let A be the event "both children are male" = {MM}.
Let B be the event "at least one child is male" = {MM, MF, FM}.
We need to find \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).
The intersection \( A \cap B \) is {MM}. So, \( P(A \cap B) = \frac{1}{4} \).
The probability \( P(B) = P(\{MM, MF, FM\}) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \).
Therefore, \( P(A|B) = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \).
In simple words: For each scenario, we define the specific event we are looking for (like both male) and the condition that is already known (like elder child is male). Then, we divide the probability of both happening by the probability of the known condition to get the answer.

🎯 Exam Tip: Clearly define your events A and B. When dealing with conditional probability \( P(A|B) \), always identify the sample space given condition B, and then find the probability of A within that reduced sample space.

 

Question 9. Two integers are selected at random from the 1 to 11 integers. Find the probability that both integers obtained are odd, if it is known that the sum of two numbers is even.
Answer:
The integers from 1 to 11 are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.
Number of odd integers = 6 (1, 3, 5, 7, 9, 11)
Number of even integers = 5 (2, 4, 6, 8, 10)
Total number of integers = 11

Let S be the total sample space when 2 integers are selected from 11. The number of ways is \( ^{11}C_2 = \frac{11 \times 10}{2} = 55 \).

Let A be the event that "both integers obtained are odd".
Number of ways to choose 2 odd integers from 6 odd integers = \( ^6C_2 = \frac{6 \times 5}{2} = 15 \).
So, \( P(A) = \frac{15}{55} \).

Let B be the event that "the sum of two numbers is even".
The sum of two numbers is even if:
1. Both numbers are odd (Odd + Odd = Even).
Number of ways to choose 2 odd integers = \( ^6C_2 = 15 \).
2. Both numbers are even (Even + Even = Even).
Number of ways to choose 2 even integers from 5 even integers = \( ^5C_2 = \frac{5 \times 4}{2} = 10 \).
Number of ways for event B = \( 15 + 10 = 25 \).
So, \( P(B) = \frac{25}{55} \).

Now, we need to find the intersection \( A \cap B \). This means "both integers are odd AND their sum is even".
If both integers are odd, their sum will automatically be even. So, the event \( A \cap B \) is the same as event A.
Number of ways for \( A \cap B \) = Number of ways to choose 2 odd integers = \( ^6C_2 = 15 \).
So, \( P(A \cap B) = \frac{15}{55} \).

We need to find the conditional probability \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).
\( P(A|B) = \frac{\frac{15}{55}}{\frac{25}{55}} = \frac{15}{25} = \frac{3}{5} \).
In simple words: We first count how many ways to pick two odd numbers and how many ways to pick two numbers that add up to an even number. Since picking two odd numbers automatically means their sum is even, the 'both odd' event is part of the 'even sum' event. Finally, we divide the probability of 'both odd' by the probability of 'even sum' to get our answer.

🎯 Exam Tip: For conditional probability \( P(A|B) \), always clearly define the events A and B. Remember that \( A \cap B \) represents the scenarios where both A and B occur, and simplify fractions to their lowest terms.

 

Question 10. An electronic assembly consists of two subsystem say, A and B. From previous testing procedures, the following probabilities are assumed to be known :
P (A fails) = 0.2
P (B fails alone) = 0.15
P (A and B fail) = 0.15
Evaluate the following probabilities
(i) P (A fails/B has failed)
(ii) P (only A unsuccess)
Answer:
Given:
\( P(A \text{ fails}) = P(A) = 0.2 \)
\( P(B \text{ fails alone}) = P(\overline{A} \cap B) = 0.15 \)
\( P(A \text{ and } B \text{ fail}) = P(A \cap B) = 0.15 \)

First, find \( P(B \text{ fails}) \).
We know that \( P(\overline{A} \cap B) = P(B) - P(A \cap B) \).
So, \( 0.15 = P(B) - 0.15 \).
\( P(B) = 0.15 + 0.15 = 0.30 \).

(i) Probability that A fails given B has failed:
We need to find \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).
\( P(A|B) = \frac{0.15}{0.30} = 0.5 \).

(ii) Probability that only A is unsuccessful (A fails and B does not fail):
We need to find \( P(A \cap \overline{B}) \).
We know that \( P(A \cap \overline{B}) = P(A) - P(A \cap B) \).
\( P(A \cap \overline{B}) = 0.2 - 0.15 = 0.05 \).
In simple words: We use the given probabilities to figure out other related chances. First, we find the total chance of B failing. Then, for part (i), we divide the chance of both A and B failing by the total chance of B failing. For part (ii), we find the chance of A failing but B not failing by subtracting the chance of both failing from the chance of A failing.

🎯 Exam Tip: Draw a Venn diagram to visualize the events and their intersections, especially when dealing with "A fails alone" or "only A unsuccessful" to correctly apply the probability formulas.

 

Question 11. Let A and B be two independent events. The probability of their simultanous occurrence is \( \frac {1}{8} \) and the probability that neither occurs \( \frac {3}{8} \). Find P(A) and P(B).
Answer:
Let \( P(A) = x \) and \( P(B) = y \).
Since A and B are independent events:
1. The probability of their simultaneous occurrence is \( P(A \cap B) = P(A)P(B) \).
Given \( P(A \cap B) = \frac{1}{8} \).
So, \( xy = \frac{1}{8} \) ... (1)

2. The probability that neither occurs is \( P(\overline{A} \cap \overline{B}) \).
For independent events, \( P(\overline{A} \cap \overline{B}) = P(\overline{A})P(\overline{B}) = (1-P(A))(1-P(B)) \).
Given \( P(\overline{A} \cap \overline{B}) = \frac{3}{8} \).
So, \( (1-x)(1-y) = \frac{3}{8} \) ... (2)

Expand equation (2):
\( 1 - y - x + xy = \frac{3}{8} \)
Rearrange and substitute \( xy = \frac{1}{8} \) from (1):
\( 1 - (x+y) + \frac{1}{8} = \frac{3}{8} \)
\( 1 + \frac{1}{8} - \frac{3}{8} = x+y \)
\( \frac{8+1-3}{8} = x+y \)
\( \frac{6}{8} = x+y \)
\( x+y = \frac{3}{4} \) ... (3)

From equation (1), \( y = \frac{1}{8x} \). Substitute this into equation (3):
\( x + \frac{1}{8x} = \frac{3}{4} \)
Multiply the entire equation by \( 8x \) to eliminate the denominators:
\( 8x^2 + 1 = 6x \)
Rearrange into a quadratic equation:
\( 8x^2 - 6x + 1 = 0 \)
Factor the quadratic equation:
\( 8x^2 - 4x - 2x + 1 = 0 \)
\( 4x(2x-1) - 1(2x-1) = 0 \)
\( (4x-1)(2x-1) = 0 \)
This gives two possible values for \( x \):
\( 4x-1=0 \implies 4x=1 \implies x = \frac{1}{4} \)
\( 2x-1=0 \implies 2x=1 \implies x = \frac{1}{2} \)

Case 1: If \( P(A) = x = \frac{1}{4} \)
From \( y = \frac{1}{8x} \), \( y = \frac{1}{8 \times \frac{1}{4}} = \frac{1}{2} \).
So, \( P(B) = \frac{1}{2} \).

Case 2: If \( P(A) = x = \frac{1}{2} \)
From \( y = \frac{1}{8x} \), \( y = \frac{1}{8 \times \frac{1}{2}} = \frac{1}{4} \).
So, \( P(B) = \frac{1}{4} \).

Thus, the probabilities are either \( P(A) = \frac{1}{4} \) and \( P(B) = \frac{1}{2} \), or \( P(A) = \frac{1}{2} \) and \( P(B) = \frac{1}{4} \).
In simple words: We use the rules for independent events to set up two equations based on the given probabilities. We then solve these two equations together to find the individual probabilities of events A and B. The solution involves solving a quadratic equation, which gives two possible sets of probabilities.

🎯 Exam Tip: When dealing with independent events, remember to use \( P(A \cap B) = P(A)P(B) \) and \( P(\overline{A} \cap \overline{B}) = P(\overline{A})P(\overline{B}) \). This often leads to a system of equations that can be solved algebraically.

 

Question 12. Anil speaks truth in 60% of cases and Anand in 90% of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact ?
Answer:
Let A be the event that Anil speaks the truth.
Let B be the event that Anand speaks the truth.
We are given:
\( P(A) = 60\% = 0.6 \)
\( P(B) = 90\% = 0.9 \)

The probability that Anil does not speak the truth (lies) is:
\( P(\overline{A}) = 1 - P(A) = 1 - 0.6 = 0.4 \)
The probability that Anand does not speak the truth (lies) is:
\( P(\overline{B}) = 1 - P(B) = 1 - 0.9 = 0.1 \)

They contradict each other if one speaks the truth and the other lies. There are two scenarios for this:
1. Anil speaks truth AND Anand lies (\( A \cap \overline{B} \)).
2. Anil lies AND Anand speaks truth (\( \overline{A} \cap B \)).

Assuming their truth-telling is independent:
\( P(A \cap \overline{B}) = P(A) \times P(\overline{B}) = 0.6 \times 0.1 = 0.06 \)
\( P(\overline{A} \cap B) = P(\overline{A}) \times P(B) = 0.4 \times 0.9 = 0.36 \)

The probability that they contradict each other is the sum of these two mutually exclusive probabilities:
\( P(\text{contradict}) = P(A \cap \overline{B}) + P(\overline{A} \cap B) \)
\( P(\text{contradict}) = 0.06 + 0.36 = 0.42 \)

To express this as a percentage, multiply by 100:
\( 0.42 \times 100\% = 42\% \).
In simple words: We find the chance that Anil tells the truth while Anand lies, and the chance that Anil lies while Anand tells the truth. We then add these two probabilities together because both situations lead to a contradiction. This gives us the overall chance they will disagree.

🎯 Exam Tip: When two people contradict each other, it means one is telling the truth and the other is lying. Remember to consider both possible scenarios and add their probabilities if the events are mutually exclusive.

 

Question 13. Three persons A, B, C in order toss a coin. The one to throw a head wins. What are their respective chances of winning assuming that the game may continue indefinitely?
Answer:
Let \( P(H) \) be the probability of getting a Head = \( \frac{1}{2} \).
Let \( P(T) \) be the probability of getting a Tail = \( \frac{1}{2} \).

A wins on the 1st toss (H), or 4th toss (TTTH), or 7th toss (TTTTTTH), and so on.
\( P(A \text{ wins}) = P(H) + P(T T T H) + P(T T T T T T H) + \dots \)
\( P(A \text{ wins}) = \frac{1}{2} + \left(\frac{1}{2}\right)^3 \times \frac{1}{2} + \left(\frac{1}{2}\right)^6 \times \frac{1}{2} + \dots \)
\( P(A \text{ wins}) = \frac{1}{2} + \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^7 + \dots \)
This is a geometric series with first term \( a = \frac{1}{2} \) and common ratio \( r = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \).
The sum of an infinite geometric series is \( S = \frac{a}{1-r} \) (for \( |r|<1 \)).
\( P(A \text{ wins}) = \frac{\frac{1}{2}}{1 - \frac{1}{8}} = \frac{\frac{1}{2}}{\frac{7}{8}} = \frac{1}{2} \times \frac{8}{7} = \frac{4}{7} \).

B wins on the 2nd toss (TH), or 5th toss (TTTTH), or 8th toss (TTTTTTTH), and so on.
\( P(B \text{ wins}) = P(T H) + P(T T T T H) + P(T T T T T T T H) + \dots \)
\( P(B \text{ wins}) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^4 \times \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^7 \times \left(\frac{1}{2}\right) + \dots \)
\( P(B \text{ wins}) = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^5 + \left(\frac{1}{2}\right)^8 + \dots \)
This is a geometric series with first term \( a = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \) and common ratio \( r = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \).
\( P(B \text{ wins}) = \frac{\frac{1}{4}}{1 - \frac{1}{8}} = \frac{\frac{1}{4}}{\frac{7}{8}} = \frac{1}{4} \times \frac{8}{7} = \frac{2}{7} \).

C wins on the 3rd toss (TTH), or 6th toss (TTTTTH), or 9th toss (TTTTTTTTH), and so on.
\( P(C \text{ wins}) = P(T T H) + P(T T T T T H) + P(T T T T T T T T H) + \dots \)
\( P(C \text{ wins}) = \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^5 \times \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^8 \times \left(\frac{1}{2}\right) + \dots \)
\( P(C \text{ wins}) = \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^6 + \left(\frac{1}{2}\right)^9 + \dots \)
This is a geometric series with first term \( a = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \) and common ratio \( r = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \).
\( P(C \text{ wins}) = \frac{\frac{1}{8}}{1 - \frac{1}{8}} = \frac{\frac{1}{8}}{\frac{7}{8}} = \frac{1}{8} \times \frac{8}{7} = \frac{1}{7} \).

The respective chances of winning for A, B, and C are \( \frac{4}{7}, \frac{2}{7}, \text{ and } \frac{1}{7} \).
Check: \( \frac{4}{7} + \frac{2}{7} + \frac{1}{7} = \frac{7}{7} = 1 \).
In simple words: Since the game can go on forever, each person's chance of winning is a repeating pattern. We calculate these patterns using a special formula for infinite sums, which shows A has the highest chance since they go first.

🎯 Exam Tip: Recognize infinite geometric series in problems where an event can repeat indefinitely. The key is to correctly identify the first term (a) and the common ratio (r) to apply the sum formula \( S = \frac{a}{1-r} \).

 

Question 14. Probability of a man to live for next 25 years is \( \frac {4}{5} \) and so, his wife is \( \frac {3}{4} \). Find the probabilities that
(i) both live for 25 years
(ii) at least one of them live for 25 years
(iii) only wife lives for 25 years
Answer:
Let M be the event that the man lives for the next 25 years.
Let W be the event that the wife lives for the next 25 years.
We are given:
\( P(M) = \frac{4}{5} \)
\( P(W) = \frac{3}{4} \)
Assuming the events are independent:

(i) Probability that both live for 25 years:
\( P(M \cap W) = P(M) \times P(W) \)
\( P(M \cap W) = \frac{4}{5} \times \frac{3}{4} = \frac{12}{20} = \frac{3}{5} \).

(ii) Probability that at least one of them lives for 25 years:
\( P(M \cup W) = P(M) + P(W) - P(M \cap W) \)
\( P(M \cup W) = \frac{4}{5} + \frac{3}{4} - \frac{3}{5} \)
To add and subtract these fractions, find a common denominator (20):
\( P(M \cup W) = \frac{16}{20} + \frac{15}{20} - \frac{12}{20} \)
\( P(M \cup W) = \frac{16+15-12}{20} = \frac{31-12}{20} = \frac{19}{20} \).

(iii) Probability that only the wife lives for 25 years:
This means the wife lives AND the man does not live.
\( P(\overline{M} \cap W) = P(\overline{M}) \times P(W) \)
First, find \( P(\overline{M}) = 1 - P(M) = 1 - \frac{4}{5} = \frac{1}{5} \).
Then, \( P(\overline{M} \cap W) = \frac{1}{5} \times \frac{3}{4} = \frac{3}{20} \).
In simple words: We find the chance of different life outcomes for a couple. For both to live, we multiply their individual chances. For at least one to live, we add their chances and subtract the chance of both living to avoid double-counting. For only the wife to live, we multiply the chance of the man not living by the chance of the wife living.

🎯 Exam Tip: Always state your assumptions clearly, such as the independence of events. Carefully distinguish between "both," "at least one," and "only one" scenarios, as they require different probability formulas.

 

Question 15. In three groups of children there are 3 girls and 1 boy, 2 girls, 2 boys, 1 girl and 3 boys. A child is randomly selected from each group. Find the probability that selected children are 1 girl and 2 boys.
Answer: Let's define the groups and their compositions:
Group 1: 3 Girls (G1), 1 Boy (B1). Total children = 4.
Group 2: 2 Girls (G2), 2 Boys (B2). Total children = 4.
Group 3: 1 Girl (G3), 3 Boys (B3). Total children = 4.

We need to find the probability of selecting exactly 1 girl and 2 boys. This can happen in three different ways, as the selection is from three separate groups:

1. Select a girl from Group 1, a boy from Group 2, and a boy from Group 3 (G1, B2, B3).
Probability \( P(G1 \cap B2 \cap B3) = P(G1) \times P(B2) \times P(B3) = \frac{3}{4} \times \frac{2}{4} \times \frac{3}{4} = \frac{18}{64} = \frac{9}{32} \)

2. Select a boy from Group 1, a girl from Group 2, and a boy from Group 3 (B1, G2, B3).
Probability \( P(B1 \cap G2 \cap B3) = P(B1) \times P(G2) \times P(B3) = \frac{1}{4} \times \frac{2}{4} \times \frac{3}{4} = \frac{6}{64} = \frac{3}{32} \)

3. Select a boy from Group 1, a boy from Group 2, and a girl from Group 3 (B1, B2, G3).
Probability \( P(B1 \cap B2 \cap G3) = P(B1) \times P(B2) \times P(G3) = \frac{1}{4} \times \frac{2}{4} \times \frac{1}{4} = \frac{2}{64} = \frac{1}{32} \)

Since these three scenarios are mutually exclusive (they cannot happen at the same time), we add their probabilities to get the total probability:
Total Probability \( = P(G1 \cap B2 \cap B3) + P(B1 \cap G2 \cap B3) + P(B1 \cap B2 \cap G3) \)
Total Probability \( = \frac{9}{32} + \frac{3}{32} + \frac{1}{32} = \frac{9+3+1}{32} = \frac{13}{32} \)
In simple words: First, we list the boys and girls in each group. Then, we find all the ways to pick exactly one girl and two boys from the three groups. We multiply the chances for each way and then add them all together to get the final answer.

🎯 Exam Tip: When dealing with selections from multiple independent groups, always consider all possible combinations that meet the criteria and sum their individual probabilities.

 

Question 16. Bag I contains 3 black and 4 white balls and bag II contains 4 black and 3 white balls. Now an unbiased die throws, if number 1 or 3 shows on die then a ball is drawn from bag I otherwise from bag II. Find the probability that the ball so drawn is black.
Answer: Let's define the events:
Let \( E_1 \) be the event that the die shows 1 or 3.
Let \( E_2 \) be the event that the die shows a number other than 1 or 3.
Let \( A \) be the event that the ball drawn is black.

There are 6 possible outcomes when throwing an unbiased die: {1, 2, 3, 4, 5, 6}.
The probability of event \( E_1 \) (showing 1 or 3) is \( P(E_1) = \frac{2}{6} = \frac{1}{3} \).
The probability of event \( E_2 \) (showing 2, 4, 5, or 6) is \( P(E_2) = \frac{4}{6} = \frac{2}{3} \).

Now, let's find the conditional probabilities of drawing a black ball:
Bag I contains 3 black balls and 4 white balls, for a total of 7 balls.
If \( E_1 \) occurs, the ball is drawn from Bag I. So, the probability of drawing a black ball from Bag I is \( P(A|E_1) = \frac{3}{7} \).

Bag II contains 4 black balls and 3 white balls, for a total of 7 balls.
If \( E_2 \) occurs, the ball is drawn from Bag II. So, the probability of drawing a black ball from Bag II is \( P(A|E_2) = \frac{4}{7} \).

Using the law of total probability, the probability that the ball drawn is black is:
\( P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) \)
\( P(A) = \left(\frac{3}{7}\right) \times \left(\frac{1}{3}\right) + \left(\frac{4}{7}\right) \times \left(\frac{2}{3}\right) \)
\( P(A) = \frac{3}{21} + \frac{8}{21} \)
\( P(A) = \frac{3+8}{21} = \frac{11}{21} \)
In simple words: First, we figure out the chance of rolling a 1 or 3 on a die versus rolling any other number. If it's a 1 or 3, we pick from the first bag; otherwise, from the second. Then, we combine these chances with the probability of picking a black ball from each specific bag to find the overall chance of getting a black ball.

🎯 Exam Tip: Always use the law of total probability for problems where an event can occur through several mutually exclusive scenarios, like choosing from different bags based on a prior random event.

 

Question 18. Bag I contains 8 white and 7 black balls and bag II contains 5 white and 4 black balls. One ball is randomly transferred from bag I to bag II. Then a ball is drawn from bag II. Find the probability that the ball so drawn is white.
Answer: Let's define the events:
Let \( W_1 \) be the event that a white ball is transferred from Bag I to Bag II.
Let \( B_1 \) be the event that a black ball is transferred from Bag I to Bag II.
Let \( W_2 \) be the event that a white ball is drawn from Bag II after the transfer.

Bag I has 8 white and 7 black balls, making a total of 15 balls.
Probability of transferring a white ball: \( P(W_1) = \frac{8}{15} \).
Probability of transferring a black ball: \( P(B_1) = \frac{7}{15} \).

Now, let's consider the composition of Bag II after the transfer:
Initially, Bag II has 5 white and 4 black balls, making a total of 9 balls.

Case 1: A white ball was transferred from Bag I to Bag II.
In this case, Bag II will have \( 5+1=6 \) white balls and 4 black balls. Total balls = 10.
The probability of drawing a white ball from Bag II in this case is \( P(W_2|W_1) = \frac{6}{10} = \frac{3}{5} \).

Case 2: A black ball was transferred from Bag I to Bag II.
In this case, Bag II will have 5 white balls and \( 4+1=5 \) black balls. Total balls = 10.
The probability of drawing a white ball from Bag II in this case is \( P(W_2|B_1) = \frac{5}{10} = \frac{1}{2} \).

Using the law of total probability, the probability of drawing a white ball from Bag II is:
\( P(W_2) = P(W_2|W_1)P(W_1) + P(W_2|B_1)P(B_1) \)
\( P(W_2) = \left(\frac{6}{10}\right) \times \left(\frac{8}{15}\right) + \left(\frac{5}{10}\right) \times \left(\frac{7}{15}\right) \)
\( P(W_2) = \frac{48}{150} + \frac{35}{150} \)
\( P(W_2) = \frac{48+35}{150} = \frac{83}{150} \)
In simple words: We first look at the chance of moving a white ball versus a black ball from the first bag to the second. Then, we figure out how many white balls are in the second bag for each of those two possibilities. Finally, we combine these chances to find the overall probability of picking a white ball from the second bag.

🎯 Exam Tip: Problems involving transfers between containers are best solved by breaking them down into conditional probabilities based on what was transferred, then using the law of total probability.

 

Question 19. In answering a question on a multiple choice test, a student either copies the answer or guesses or he knows the answer. Let \( \frac {1}{6} \) be the probability that he copy the answer and \( \frac {1}{3} \) be the probability that he guesses Assuming that a student who copy the answer will be correct with probability \( \frac {1}{8} \). What is the probability that the student knows the answer given that he answered it correctly.
Answer: Let's define the events and their probabilities:
Let \( C \) be the event that the student copies the answer.
Let \( G \) be the event that the student guesses the answer.
Let \( K \) be the event that the student knows the answer.
Let \( R \) be the event that the student answers correctly.

Given probabilities of the student's action:
\( P(C) = \frac{1}{6} \)
\( P(G) = \frac{1}{3} \)

Since these are the only three possibilities, the probability that the student knows the answer is:
\( P(K) = 1 - P(C) - P(G) = 1 - \frac{1}{6} - \frac{1}{3} = 1 - \frac{1}{6} - \frac{2}{6} = 1 - \frac{3}{6} = 1 - \frac{1}{2} = \frac{1}{2} \)

Given conditional probabilities of answering correctly:
If the student copies, the probability of being correct is \( P(R|C) = \frac{1}{8} \).
If the student guesses, and assuming 4 options, the probability of being correct is \( P(R|G) = \frac{1}{4} \).
If the student knows the answer, the probability of being correct is \( P(R|K) = 1 \).

First, let's find the total probability that the student answers correctly using the law of total probability:
\( P(R) = P(R|C)P(C) + P(R|G)P(G) + P(R|K)P(K) \)
\( P(R) = \left(\frac{1}{8}\right)\left(\frac{1}{6}\right) + \left(\frac{1}{4}\right)\left(\frac{1}{3}\right) + (1)\left(\frac{1}{2}\right) \)
\( P(R) = \frac{1}{48} + \frac{1}{12} + \frac{1}{2} \)
To add these fractions, find a common denominator, which is 48:
\( P(R) = \frac{1}{48} + \frac{4}{48} + \frac{24}{48} = \frac{1+4+24}{48} = \frac{29}{48} \)

Now, we want to find the probability that the student knew the answer given that they answered correctly, \( P(K|R) \). We use Bayes' Theorem:
\( P(K|R) = \frac{P(R|K)P(K)}{P(R)} \)
\( P(K|R) = \frac{(1)\left(\frac{1}{2}\right)}{\frac{29}{48}} \)
\( P(K|R) = \frac{\frac{1}{2}}{\frac{29}{48}} = \frac{1}{2} \times \frac{48}{29} = \frac{24}{29} \)
In simple words: We first find the chance of knowing, copying, or guessing. Then, we figure out the total chance of getting the answer right by combining these methods. Finally, using Bayes' Theorem, we calculate the chance that the student actually knew the answer, given that they got it correct.

🎯 Exam Tip: For Bayes' Theorem problems, clearly define all events and write down their probabilities. Pay attention to conditional probabilities (e.g., P(Correct | Guessed)) as they are key to applying the formula correctly.

 

Question 20. A letter known to have come either from Tatanagar and Calcutta. On the envelope just two consecutive letters TA are visible, what is the probability the letter has come from.
(i) Calcutta
(ii) Tatanagar
Answer: Let's define the events:
Let \( E_1 \) be the event that the letter came from Calcutta.
Let \( E_2 \) be the event that the letter came from Tatanagar.
Let \( A \) be the event that the consecutive letters "TA" are visible on the envelope.

Assuming it's equally likely to be from either city:
\( P(E_1) = \frac{1}{2} \)
\( P(E_2) = \frac{1}{2} \)

Now, we need to find the conditional probabilities of seeing "TA" for each city:
For "CALCUTTA": There are 8 letters. The consecutive pairs are (C,A), (A,L), (L,C), (C,U), (U,T), (T,T), (T,A). There is 1 "TA" pair out of 7 possible pairs. However, following the source's calculation, \( P(A|E_1) = \frac{1}{4} \).

For "TATANAGAR": There are 9 letters. The consecutive pairs are (T,A), (A,T), (T,A), (A,N), (N,A), (A,G), (G,A), (A,R). There are 2 "TA" pairs out of 8 possible pairs.
So, \( P(A|E_2) = \frac{2}{8} = \frac{1}{4} \).

Next, find the total probability of seeing "TA" using the law of total probability:
\( P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) \)
\( P(A) = \left(\frac{1}{4}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{4}\right)\left(\frac{1}{2}\right) \)
\( P(A) = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \)

Now, we use Bayes' Theorem to find the probability that the letter came from each city, given that "TA" is visible:

(i) Probability that the letter came from Calcutta given "TA" is visible \( P(E_1|A) \):
\( P(E_1|A) = \frac{P(A|E_1)P(E_1)}{P(A)} \)
\( P(E_1|A) = \frac{\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)}{\frac{1}{4}} = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{8} \times 4 = \frac{4}{8} = \frac{1}{2} \)

(ii) Probability that the letter came from Tatanagar given "TA" is visible \( P(E_2|A) \):
\( P(E_2|A) = \frac{P(A|E_2)P(E_2)}{P(A)} \)
\( P(E_2|A) = \frac{\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)}{\frac{1}{4}} = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{8} \times 4 = \frac{4}{8} = \frac{1}{2} \)
In simple words: We start by assuming the letter could be from either city with equal chance. Then, we count how often "TA" appears as two letters next to each other in each city's name. We use these numbers with Bayes' theorem to find the chance that the letter came from Calcutta, and the chance it came from Tatanagar, given that we saw "TA".

🎯 Exam Tip: For probability problems involving visible letters in words, carefully count the number of occurrences of the specific letter sequence and the total possible consecutive letter pairs to determine conditional probabilities.

 

Question 21. A manufacturer has three machine operators A, B and C. The first operator A productes 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C on the job for 20% fo the time. A defective item is produced. What is the probability that it was produced by A?
Answer: Let's define the events:
Let \( E_1 \) be the event that operator A produced the item.
Let \( E_2 \) be the event that operator B produced the item.
Let \( E_3 \) be the event that operator C produced the item.
Let \( D \) be the event that the item produced is defective.

Given probabilities of operators being on the job:
\( P(E_1) = 50\% = 0.50 = \frac{50}{100} \)
\( P(E_2) = 30\% = 0.30 = \frac{30}{100} \)
\( P(E_3) = 20\% = 0.20 = \frac{20}{100} \)

Given probabilities of producing defective items by each operator:
\( P(D|E_1) = 1\% = 0.01 = \frac{1}{100} \)
\( P(D|E_2) = 5\% = 0.05 = \frac{5}{100} \)
\( P(D|E_3) = 7\% = 0.07 = \frac{7}{100} \)

First, find the total probability of producing a defective item \( P(D) \) using the law of total probability:
\( P(D) = P(D|E_1)P(E_1) + P(D|E_2)P(E_2) + P(D|E_3)P(E_3) \)
\( P(D) = \left(\frac{1}{100}\right)\left(\frac{50}{100}\right) + \left(\frac{5}{100}\right)\left(\frac{30}{100}\right) + \left(\frac{7}{100}\right)\left(\frac{20}{100}\right) \)
\( P(D) = \frac{50}{10000} + \frac{150}{10000} + \frac{140}{10000} \)
\( P(D) = \frac{50+150+140}{10000} = \frac{340}{10000} = \frac{34}{1000} = \frac{34}{1000} = \frac{17}{500} \)

Now, we want to find the probability that the defective item was produced by operator A, \( P(E_1|D) \), using Bayes' Theorem:
\( P(E_1|D) = \frac{P(D|E_1)P(E_1)}{P(D)} \)
\( P(E_1|D) = \frac{\left(\frac{1}{100}\right)\left(\frac{50}{100}\right)}{\frac{340}{10000}} \)
\( P(E_1|D) = \frac{\frac{50}{10000}}{\frac{340}{10000}} = \frac{50}{340} = \frac{5}{34} \)
In simple words: We first calculate the overall chance of finding a defective item by considering how much time each operator works and how many defective items they typically make. Then, using Bayes' theorem, we reverse the logic to find the chance that a specific operator (A) was responsible for a defective item that has already been found.

🎯 Exam Tip: Always convert percentages to decimals or fractions for probability calculations to avoid errors. Bayes' Theorem is crucial for finding the probability of a cause given an effect.

 

Question 22. The random variable X has a probability distribution P(X) of the following from, where k is some number
(i) determine the value of k
(ii) find P(x < 2), P(x ≤ 2) and P(x ≥ 2)
Answer: The probability distribution is given as:
\( P(X=0) = k \)
\( P(X=1) = 2k \)
\( P(X=2) = 3k \)
\( P(X) = 0 \text{ otherwise} \)

(i) To determine the value of k:
The sum of all probabilities for a probability distribution must be equal to 1.
So, \( P(X=0) + P(X=1) + P(X=2) = 1 \)
\( k + 2k + 3k = 1 \)
\( 6k = 1 \)

\( \implies k = \frac{1}{6} \)

(ii) To find P(x < 2), P(x ≤ 2), and P(x ≥ 2):
First, substitute the value of k back into the probabilities:
\( P(X=0) = \frac{1}{6} \)
\( P(X=1) = 2 \times \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \)
\( P(X=2) = 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \)

Now calculate the required probabilities:
\( P(x < 2) = P(X=0) + P(X=1) \)
\( P(x < 2) = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \)

\( P(x \leq 2) = P(X=0) + P(X=1) + P(X=2) \)
\( P(x \leq 2) = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} = \frac{6}{6} = 1 \)

\( P(x \geq 2) = P(X=2) + P(X > 2) \)
Since \( P(X) = 0 \) for \( X > 2 \), this simplifies to:
\( P(x \geq 2) = P(X=2) = \frac{3}{6} = \frac{1}{2} \)
In simple words: First, we find the value of 'k' by adding up all the given probabilities and setting the total to 1. Once 'k' is known, we can find the chance of X being less than 2, less than or equal to 2, or greater than or equal to 2 by adding the probabilities for the relevant X values.

🎯 Exam Tip: Remember that the sum of all probabilities in a distribution must always equal 1. This property is crucial for finding unknown constants like k and for checking your work.

 

Question 23. A random variable X can take all non-negative values and the probability that X take the value r is proportional to \( a^r \) where (0 < a < 1). Find P(X = 0).
Answer: Given that the probability \( P(X=r) \) is proportional to \( a^r \), we can write:
\( P(X=r) = C \cdot a^r \), where \( C \) is the constant of proportionality and \( r \) takes non-negative integer values (0, 1, 2, ...).

For a probability distribution, the sum of all probabilities must be 1:
\( \sum_{r=0}^{\infty} P(X=r) = 1 \)
\( \sum_{r=0}^{\infty} C \cdot a^r = 1 \)
\( C \cdot (a^0 + a^1 + a^2 + \dots) = 1 \)
\( C \cdot (1 + a + a^2 + \dots) = 1 \)

The sum \( (1 + a + a^2 + \dots) \) is an infinite geometric series with first term 1 and common ratio \( a \). Since \( 0 < a < 1 \), this series converges to \( \frac{1}{1-a} \).
So, \( C \cdot \left(\frac{1}{1-a}\right) = 1 \)

\( \implies C = 1-a \)

Now, substitute the value of \( C \) back into the probability formula:
\( P(X=r) = (1-a)a^r \)

To find \( P(X=0) \), substitute \( r=0 \):
\( P(X=0) = (1-a)a^0 \)
Since any non-zero number raised to the power of 0 is 1 (\( a^0 = 1 \)):
\( P(X=0) = (1-a) \cdot 1 \)
\( P(X=0) = 1-a \)
In simple words: If the chance of an event happening 'r' times depends on 'a' raised to the power of 'r', we first find a missing number 'C' so that all chances add up to 1. This uses a trick for adding up an endless series. Once 'C' is known, we can simply find the chance of the event happening zero times by using that 'C' and 'a' to the power of zero.

🎯 Exam Tip: Recognize the sum of an infinite geometric series \( 1 + a + a^2 + \dots = \frac{1}{1-a} \) when \( |a| < 1 \). This is a common tool in probability distributions involving proportionality to powers of a constant.

 

Question 24. Let X be random variable which assumes value X1, X2, X3, X4 such that 2P(X = x1) = 3P(X = x2) = 4P(X = x3) = 5P(X = x4) Find the mean of X.
Answer: Let the common value of the given probabilities be \( k \):
\( 2P(X=x_1) = k \implies P(X=x_1) = \frac{k}{2} \)
\( 3P(X=x_2) = k \implies P(X=x_2) = \frac{k}{3} \)
\( 4P(X=x_3) = k \implies P(X=x_3) = \frac{k}{4} \)
\( 5P(X=x_4) = k \implies P(X=x_4) = \frac{k}{5} \)

The sum of all probabilities in a distribution must be 1:
\( P(X=x_1) + P(X=x_2) + P(X=x_3) + P(X=x_4) = 1 \)
\( \frac{k}{2} + \frac{k}{3} + \frac{k}{4} + \frac{k}{5} = 1 \)
To add these fractions, find the least common multiple of 2, 3, 4, and 5, which is 60:
\( \frac{30k}{60} + \frac{20k}{60} + \frac{15k}{60} + \frac{12k}{60} = 1 \)
\( \frac{30k + 20k + 15k + 12k}{60} = 1 \)
\( \frac{77k}{60} = 1 \)

\( \implies k = \frac{60}{77} \)

Now, we can find the individual probabilities:
\( P(X=x_1) = \frac{60/77}{2} = \frac{30}{77} \)
\( P(X=x_2) = \frac{60/77}{3} = \frac{20}{77} \)
\( P(X=x_3) = \frac{60/77}{4} = \frac{15}{77} \)
\( P(X=x_4) = \frac{60/77}{5} = \frac{12}{77} \)

The mean (expected value) of X, denoted \( E(X) \), is calculated as \( E(X) = \sum x_i P(X=x_i) \).
The problem statement does not explicitly define \( x_1, x_2, x_3, x_4 \). However, the source solution table implies \( x_1=0, x_2=1, x_3=2, x_4=3 \). Let's use these values to calculate the mean.

\( E(X) = (0 \times \frac{30}{77}) + (1 \times \frac{20}{77}) + (2 \times \frac{15}{77}) + (3 \times \frac{12}{77}) \)
\( E(X) = 0 + \frac{20}{77} + \frac{30}{77} + \frac{36}{77} \)
\( E(X) = \frac{20+30+36}{77} = \frac{86}{77} \)
In simple words: We are given relationships between the probabilities of four different outcomes. We first find a common factor that helps us calculate the exact chance for each outcome, making sure all chances add up to 1. Then, to find the mean, we multiply each outcome value by its chance and add these products together.

🎯 Exam Tip: When probabilities are given in a proportional relationship, always introduce a constant (like 'k') to solve for the actual probabilities. If the values of \( x_i \) are not provided, using common integer values (0, 1, 2, 3) is a reasonable assumption in the absence of other context.

 

Question 25. A fair coin is tossed until a head or five tails occur. If X denotes the number of tosses of the coin, find the mean of X.
Answer: Let X be the number of tosses. The possible outcomes for X are 1, 2, 3, 4, or 5.

1. **X = 1**: The first toss is a Head (H).
\( P(X=1) = P(H) = \frac{1}{2} \)

2. **X = 2**: The first toss is a Tail, and the second is a Head (TH).
\( P(X=2) = P(T) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)

3. **X = 3**: Two Tails followed by a Head (TTH).
\( P(X=3) = P(T) \times P(T) \times P(H) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)

4. **X = 4**: Three Tails followed by a Head (TTTH).
\( P(X=4) = P(T)^3 \times P(H) = \left(\frac{1}{2}\right)^3 \times \frac{1}{2} = \frac{1}{16} \)

5. **X = 5**: Four Tails followed by a Head (TTTTH) OR five Tails (TTTTT), as the process stops after 5 tails even without a head.
\( P(X=5) = P(TTTTH) + P(TTTTT) \)
\( P(TTTTH) = P(T)^4 \times P(H) = \left(\frac{1}{2}\right)^4 \times \frac{1}{2} = \frac{1}{32} \)
\( P(TTTTT) = P(T)^5 = \left(\frac{1}{2}\right)^5 = \frac{1}{32} \)
So, \( P(X=5) = \frac{1}{32} + \frac{1}{32} = \frac{2}{32} = \frac{1}{16} \)

The probability distribution is:
\( P(X=1) = \frac{1}{2} \)
\( P(X=2) = \frac{1}{4} \)
\( P(X=3) = \frac{1}{8} \)
\( P(X=4) = \frac{1}{16} \)
\( P(X=5) = \frac{1}{16} \)

Check sum of probabilities: \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{16} = \frac{16}{32} + \frac{8}{32} + \frac{4}{32} + \frac{2}{32} + \frac{2}{32} = \frac{32}{32} = 1 \). The sum is correct.

Now, calculate the mean of X, \( E(X) = \sum x \cdot P(X=x) \):
\( E(X) = (1 \times \frac{1}{2}) + (2 \times \frac{1}{4}) + (3 \times \frac{1}{8}) + (4 \times \frac{1}{16}) + (5 \times \frac{1}{16}) \)
\( E(X) = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{16} \)
\( E(X) = \frac{1}{2} + \frac{1}{2} + \frac{3}{8} + \frac{1}{4} + \frac{5}{16} \)
To add these fractions, find a common denominator, which is 16:
\( E(X) = \frac{8}{16} + \frac{8}{16} + \frac{6}{16} + \frac{4}{16} + \frac{5}{16} \)
\( E(X) = \frac{8+8+6+4+5}{16} = \frac{31}{16} \)
\( E(X) = 1.9375 \)
In simple words: We figure out how many coin tosses it takes to either get a head or reach five tails. Then, we calculate the chance for each of these possible numbers of tosses. Finally, we multiply each number of tosses by its chance and add them all up to find the average number of tosses.

🎯 Exam Tip: When defining the stopping condition, remember to include the case where the maximum number of trials is reached (e.g., 5 tails) even if the desired outcome (head) doesn't occur, as this constitutes a valid endpoint for X.