OP Malhotra Class 12 Maths Solutions Chapter 18 Probability Exercise 18 (C)

S Chand Class 12 ICSE Maths Solutions Chapter 18 Probability Ex 18(c)

Question 1. A coin is tossed thrice and all eight outcomes are equally likely, E : “the first throw results in head” F: “the last throw results in tail”. Prove that events E and F are independent.
Answer: When a coin is tossed three times, the total possible results (sample space S) are:
\( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \)
So, the total number of outcomes is \( n(S) = 8 \). A coin toss is a simple random experiment.

Event E: "the first throw results in head"
\( E = \{HHH, HHT, HTH, HTT\} \)
Number of outcomes in E is \( n(E) = 4 \).
Probability of E is \( P(E) = \frac{n(E)}{n(S)} = \frac{4}{8} = \frac{1}{2} \).

Event F: "the last throw results in tail"
\( F = \{HHT, HTT, THT, TTT\} \)
Number of outcomes in F is \( n(F) = 4 \).
Probability of F is \( P(F) = \frac{n(F)}{n(S)} = \frac{4}{8} = \frac{1}{2} \).

Now, find the intersection of events E and F (E and F both happen):
\( E \cap F = \{HHT, HTT\} \)
Number of outcomes in \( E \cap F \) is \( n(E \cap F) = 2 \).
Probability of \( E \cap F \) is \( P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{2}{8} = \frac{1}{4} \).

For events E and F to be independent, the condition \( P(E \cap F) = P(E) \times P(F) \) must be true.
Let's check this:
\( P(E) \times P(F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).

Since \( P(E \cap F) = \frac{1}{4} \) and \( P(E) \times P(F) = \frac{1}{4} \), we can say:
\( P(E \cap F) = P(E) \times P(F) \)

Therefore, events E and F are independent.
In simple words: We listed all possible results when a coin is tossed three times. Then we found the results for the first throw being a head and the last throw being a tail. When the chance of both events happening together is the same as multiplying their individual chances, it means they don't affect each other, so they are independent.

🎯 Exam Tip: To prove independence, always clearly define your sample space, events, their individual probabilities, and the probability of their intersection. The key is to verify if \( P(A \cap B) = P(A)P(B) \).

Question 2. A card drawn from a well shuffled deck of 52 cards. The events A and B are A : getting a card of spade, B: getting an ace. Determine whether the events A and B are independent or not.
Answer: The total number of cards in a well-shuffled deck is \( n(S) = 52 \).

Event A: getting a card of spade.
There are 13 spade cards in a deck. A deck has four suits: clubs, diamonds, hearts, and spades, each with 13 cards.
So, \( n(A) = 13 \).
Probability of A is \( P(A) = \frac{n(A)}{n(S)} = \frac{13}{52} = \frac{1}{4} \).

Event B: getting an ace card.
There are 4 ace cards in a deck (Ace of spades, Ace of clubs, Ace of diamonds, Ace of hearts).
So, \( n(B) = 4 \).
Probability of B is \( P(B) = \frac{n(B)}{n(S)} = \frac{4}{52} = \frac{1}{13} \).

Event \( A \cap B \): getting an ace card of spade.
There is only one card that is both an ace and a spade (the Ace of Spades).
So, \( n(A \cap B) = 1 \).
Probability of \( A \cap B \) is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{52} \).

Now, let's check the condition for independence: \( P(A \cap B) = P(A) \times P(B) \).
Calculate \( P(A) \times P(B) \):
\( P(A) \times P(B) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52} \).

Since \( P(A \cap B) = \frac{1}{52} \) and \( P(A) \times P(B) = \frac{1}{52} \), we find that:
\( P(A \cap B) = P(A) \times P(B) \)

Therefore, events A and B are independent events.
In simple words: We checked if getting a spade and getting an ace are connected. Since the chance of getting the Ace of Spades is the same as multiplying the chance of getting any spade by the chance of getting any ace, these two events do not influence each other.

🎯 Exam Tip: Remember that in a standard deck, the Ace of Spades is the only card that satisfies both conditions simultaneously. This is crucial for calculating \( P(A \cap B) \).

Question 3. A bag contains 7 green, 4 white and 5 red balls. If four balls are drawn one by one with replacement, what is the probability that none is red.
Answer: Given number of green balls = 7.
Number of white balls = 4.
Number of red balls = 5.
Total number of balls in the bag is \( 7 + 4 + 5 = 16 \). This bag has a good mix of colors.

We want to find the probability that none of the balls drawn are red. This means we are drawing non-red balls.
Number of non-red balls = Number of green balls + Number of white balls = \( 7 + 4 = 11 \).

The probability of drawing a non-red ball in a single draw is:
\( P(\text{non-red}) = \frac{\text{Number of non-red balls}}{\text{Total number of balls}} = \frac{11}{16} \).

Since four balls are drawn one by one **with replacement**, each draw is an independent event. The probability of drawing a non-red ball remains the same for each draw.

The required probability that none of the four balls drawn are red is the product of the probabilities of drawing a non-red ball in each of the four draws:
Required probability \( = P(\text{non-red in 1st draw}) \times P(\text{non-red in 2nd draw}) \times P(\text{non-red in 3rd draw}) \times P(\text{non-red in 4th draw}) \)
\( = \frac{11}{16} \times \frac{11}{16} \times \frac{11}{16} \times \frac{11}{16} = \left(\frac{11}{16}\right)^4 \).
In simple words: First, count all the balls that are not red. Then, find the chance of picking one of these non-red balls. Since you put the ball back each time, the chance is the same for all four picks. Just multiply that chance by itself four times.

🎯 Exam Tip: For "with replacement" problems, each event is independent, so you can simply multiply the individual probabilities. Always be careful to identify if it's "with replacement" or "without replacement" as it changes the probabilities for subsequent draws.

Question 4. (i) A and B are two independent events such that P (A ∪ B) 0.5 and P (A) = 0.2, find P(B).
(ii) If A and B are two independent events such that P (A) = 0.3 and P (B) = 0.6, then find P(A or B).
Answer:
(i) Given that A and B are independent events, we know that \( P(A \cap B) = P(A) \times P(B) \).
We are given \( P(A \cup B) = 0.5 \) and \( P(A) = 0.2 \). We need to find \( P(B) \).
The general formula for the union of two events is: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
Substitute the independence condition into this formula:
\( P(A \cup B) = P(A) + P(B) - P(A) \times P(B) \).

Now, plug in the given values:
\( 0.5 = 0.2 + P(B) - 0.2 \times P(B) \)
\( 0.5 - 0.2 = P(B) (1 - 0.2) \)
\( 0.3 = P(B) (0.8) \)
\( P(B) = \frac{0.3}{0.8} = \frac{3}{8} \).

(ii) Given that A and B are independent events, we again use \( P(A \cap B) = P(A) \times P(B) \).
We are given \( P(A) = 0.3 \) and \( P(B) = 0.6 \). We need to find \( P(A \text{ or } B) \), which is \( P(A \cup B) \).
Using the formula for the union of two independent events:
\( P(A \cup B) = P(A) + P(B) - P(A) \times P(B) \).

Plug in the given values:
\( P(A \cup B) = 0.3 + 0.6 - (0.3 \times 0.6) \)
\( P(A \cup B) = 0.9 - 0.18 \)
\( P(A \cup B) = 0.72 \).
In simple words: For part (i), we used a special formula for independent events that connects the chance of A or B happening, the chance of A, and the chance of B. We then solved it to find the unknown chance of B. For part (ii), we used the same special formula with the given chances of A and B to find the chance of A or B happening. Understanding event independence simplifies many probability calculations.

🎯 Exam Tip: Always remember the formula for the union of two independent events: \( P(A \cup B) = P(A) + P(B) - P(A)P(B) \). This avoids calculating \( P(A \cap B) \) separately, which is only valid for independent events.

Question 5. Tickets, numbered from 1 to 19. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both the tickets will show even number.
Answer: The total number of tickets in the bag is 19. The tickets are numbered from 1 to 19.

First, let's list the even numbers between 1 and 19:
\( \{2, 4, 6, 8, 10, 12, 14, 16, 18\} \).
There are 9 even numbers among the 19 tickets. Knowing the types of numbers helps with probability.

Probability of drawing an even number on the first draw:
\( P(\text{1st even}) = \frac{\text{Number of even tickets}}{\text{Total number of tickets}} = \frac{9}{19} \).

Now, a second ticket is drawn **without replacement**. This means the first ticket is not put back.
After drawing one even ticket, the total number of tickets left in the bag is \( 19 - 1 = 18 \).
Also, the number of even tickets remaining is \( 9 - 1 = 8 \).

Probability of drawing an even number on the second draw (given the first was even and not replaced):
\( P(\text{2nd even | 1st even}) = \frac{\text{Number of remaining even tickets}}{\text{Total remaining tickets}} = \frac{8}{18} \).

The required probability that both tickets drawn are even is the product of these two probabilities:
Required probability \( = P(\text{1st even}) \times P(\text{2nd even | 1st even}) \)
\( = \frac{9}{19} \times \frac{8}{18} \).
Simplify the fraction:
\( = \frac{9}{19} \times \frac{4}{9} \)
\( = \frac{4}{19} \).
In simple words: First, count how many even numbers there are out of 19 tickets. That's your chance for the first pick. Then, because you don't put the ticket back, you have one less ticket and one less even number, so the chance for the second pick changes. Multiply these two chances together to get the final answer.

🎯 Exam Tip: For "without replacement" problems, remember that both the numerator (number of favorable outcomes) and the denominator (total outcomes) decrease by one for each successive draw.

Question 6. An unbiased die is tossed twice. Find the probability of getting a 4,5,6 on the first toss and a 1, 2,3, or 4 on the second toss.
Answer: When an unbiased die is tossed, the possible outcomes are \( \{1, 2, 3, 4, 5, 6\} \). The total number of outcomes for one toss is 6. Each toss is independent of the other.

For the first toss, we want the outcome to be a 4, 5, or 6.
Let A be the event of getting a 4, 5, or 6 on the first toss. The favorable outcomes are \( \{4, 5, 6\} \).
Number of favorable outcomes for A is \( n(A) = 3 \).
Probability of A is \( P(A) = \frac{3}{6} = \frac{1}{2} \).

For the second toss, we want the outcome to be a 1, 2, 3, or 4.
Let B be the event of getting a 1, 2, 3, or 4 on the second toss. The favorable outcomes are \( \{1, 2, 3, 4\} \).
Number of favorable outcomes for B is \( n(B) = 4 \).
Probability of B is \( P(B) = \frac{4}{6} = \frac{2}{3} \).

Since the two tosses are independent events, the probability of both events happening is the product of their individual probabilities. An unbiased die ensures fair and independent results.
Required probability \( = P(A) \times P(B) \)
\( = \frac{3}{6} \times \frac{4}{6} \)
\( = \frac{12}{36} \)
\( = \frac{1}{3} \).
In simple words: First, find the chance of getting a 4, 5, or 6 on one throw of a die. Then, find the chance of getting a 1, 2, 3, or 4 on a second throw. Since these two throws don't affect each other, just multiply their chances together to get the answer.

🎯 Exam Tip: For independent events like successive die rolls or coin tosses, always calculate the individual probabilities for each event and then multiply them to find the probability of all events occurring in sequence.

Question 7. (i) Find the probability of getting head in both trials, when a balanced coin is tossed twice.
(ii) Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability of drawing two aces.
(iii) A die is thrown three times. Getting a 5 or 6 is considered a success. Find the probability of getting (a) 3 successes; (b) exactly 2 successes; (c) at most 2 successes, (d) at least 2 successes.
(iv) A die is thrown 3 times. Getting a multiple of 3 is considered a success. Find the probability of at least 2 successes.
Answer:
(i) When a balanced coin is tossed, the probability of getting a head (H) is \( P(H) = \frac{1}{2} \), and the probability of getting a tail (T) is \( P(T) = \frac{1}{2} \). Each toss is independent.
For two trials, the outcomes are (HH, HT, TH, TT).
The probability of getting head in both trials (HH) is:
\( P(\text{HH}) = P(H) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).

(ii) A standard deck has 52 cards. There are 4 aces in the deck. Cards are drawn successively with replacement, so events are independent.
Probability of drawing an ace in the first draw = \( \frac{4}{52} = \frac{1}{13} \).
Since the card is replaced, the probability of drawing an ace in the second draw is also \( \frac{4}{52} = \frac{1}{13} \).
The required probability of drawing two aces is:
\( P(\text{two aces}) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169} \).

(iii) A die is thrown three times. Getting a 5 or 6 is a success.
Possible outcomes on a die are \( \{1, 2, 3, 4, 5, 6\} \).
Probability of success (getting 5 or 6), \( p = \frac{2}{6} = \frac{1}{3} \).
Probability of failure (not getting 5 or 6), \( q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \).
Let X be the number of successes in 3 trials. This is a binomial probability scenario, which is common in repeated independent trials.

(a) Probability of getting 3 successes (ppp):
\( P(X=3) = p \times p \times p = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \).

(b) Probability of getting exactly 2 successes (ppq, pqp, qpp):
\( P(X=2) = 3 \times p^2 \times q = 3 \times \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27} = \frac{2}{9} \).
Alternatively, using individual sequences:
\( P(\text{ppq}) = \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{27} \)
\( P(\text{pqp}) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27} \)
\( P(\text{qpp}) = \frac{2}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{2}{27} \)
Total \( = \frac{2}{27} + \frac{2}{27} + \frac{2}{27} = \frac{6}{27} = \frac{2}{9} \).

(c) Probability of getting at most 2 successes:
This means 0, 1, or 2 successes. It's easier to calculate \( 1 - P(\text{3 successes}) \).
\( P(X \le 2) = 1 - P(X=3) = 1 - \frac{1}{27} = \frac{26}{27} \).

(d) Probability of getting at least 2 successes:
This means 2 or 3 successes. It is \( P(X=2) + P(X=3) \).
\( P(X \ge 2) = P(X=2) + P(X=3) = \frac{6}{27} + \frac{1}{27} = \frac{7}{27} \).

(iv) A die is thrown 3 times. Getting a multiple of 3 is a success.
Multiples of 3 on a die are \( \{3, 6\} \).
Probability of success \( p = \frac{2}{6} = \frac{1}{3} \).
Probability of failure \( q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \).
We need the probability of at least 2 successes, which means 2 or 3 successes.
This is the same calculation as part (iii)(d) since the probabilities p and q are identical.
\( P(X \ge 2) = P(X=2) + P(X=3) \).
\( P(X=2) = 3 \times \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{6}{27} \).
\( P(X=3) = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \).
Therefore, \( P(X \ge 2) = \frac{6}{27} + \frac{1}{27} = \frac{7}{27} \).
In simple words: For coin tosses and card draws with replacement, just multiply the chances together. For die rolls with "success" defined, calculate the chance of success (p) and failure (q). Then, use these to find the chances of getting exactly 2, 3, at most 2, or at least 2 successes. It is often easier to calculate "at most" or "at least" probabilities by using the complement rule (1 minus the opposite event).

🎯 Exam Tip: When dealing with multiple trials and 'success' events, identify if it's a binomial distribution problem. Remember that 'at most n successes' means \( P(X \le n) \), and 'at least n successes' means \( P(X \ge n) \). The complement rule \( P(X \le n) = 1 - P(X > n) \) is a powerful shortcut.

Question 8. Four cards are drawn successively one after the other from a well shuffled pack of 52 cards. If the cards are not replaced, find the probability that all of them are aces.
Answer: We have a well-shuffled pack of 52 cards. We are drawing four cards one after another without replacement. This means each draw affects the next. Let's think about the deck changing after each pick.

Let E be the event that the first card drawn is an ace.
There are 4 aces in 52 cards.
\( P(E) = \frac{4}{52} \).

Let F be the event that the second card drawn is an ace (given the first was an ace and not replaced).
Now there are 51 cards left, and 3 aces remaining.
\( P(F|E) = \frac{3}{51} \).

Let G be the event that the third card drawn is an ace (given the first two were aces and not replaced).
Now there are 50 cards left, and 2 aces remaining.
\( P(G|E \cap F) = \frac{2}{50} \).

Let H be the event that the fourth card drawn is an ace (given the first three were aces and not replaced).
Now there are 49 cards left, and 1 ace remaining.
\( P(H|E \cap F \cap G) = \frac{1}{49} \).

The required probability that all four cards drawn are aces is the product of these conditional probabilities:
\( P(E \cap F \cap G \cap H) = P(E) \times P(F|E) \times P(G|E \cap F) \times P(H|E \cap F \cap G) \)
\( = \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49} \).
Let's simplify this product:
\( = \frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} \times \frac{1}{49} \)
\( = \frac{1}{13 \times 17 \times 25 \times 49} \)
\( = \frac{1}{270725} \).
In simple words: When you pick cards and don't put them back, the chances change for every new pick. For the first ace, you have 4 chances out of 52. For the second, 3 out of 51, and so on. To find the chance of getting four aces in a row, you multiply all these changing chances together.

🎯 Exam Tip: Problems involving "without replacement" are about dependent events. Each probability calculation must reflect the reduced number of total items and favorable items remaining from the previous draws.

Question 9. A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and the probability that both 4 and B are selected is at most 0.3. Is it possible that the probability of B getting selected is 0.9?
Answer: Let A be the event that candidate A is selected, and B be the event that candidate B is selected.
We are given: \( P(A) = 0.5 \).
We are also given that the probability that both A and B are selected is at most 0.3, so \( P(A \cap B) \le 0.3 \).
It is stated that A and B are independent events. This is an important piece of information for calculation.

For independent events, the probability of both A and B being selected is \( P(A \cap B) = P(A) \times P(B) \).
Using the given information: \( P(A) \times P(B) \le 0.3 \).
Substitute the value of \( P(A) \):
\( 0.5 \times P(B) \le 0.3 \).

Now, let's solve for \( P(B) \):
\( P(B) \le \frac{0.3}{0.5} \)
\( P(B) \le \frac{3}{5} \)
\( P(B) \le 0.6 \).

This calculation shows that the probability of B getting selected, \( P(B) \), cannot be more than 0.6. The selection processes are separate but restricted by the combined outcome.

The question asks if it is possible that \( P(B) = 0.9 \).
Since our calculated maximum value for \( P(B) \) is 0.6, and \( 0.9 > 0.6 \), it is not possible for the probability of B getting selected to be 0.9.
In simple words: If two events happen independently, the chance of both happening together is found by multiplying their individual chances. We know the chance for A and the highest chance for both A and B. From this, we worked out the highest possible chance for B. Since the given chance for B (0.9) is higher than this maximum possible value (0.6), it just can't be true.

🎯 Exam Tip: Always pay attention to inequality signs (at most, at least) when setting up your equations. For independent events, remember to use \( P(A \cap B) = P(A)P(B) \) to establish a constraint or range for probabilities.

Question 10. Given P (A) = 1/4, P (B/A) = 1/2 and P (A/B) = 1/4. Find, if (i) A and B are mutually exclusive, (ii) A and B are independent.
Answer: We are given the following probabilities:
\( P(A) = \frac{1}{4} \)
\( P(B/A) = \frac{1}{2} \) (probability of B given A)
\( P(A/B) = \frac{1}{4} \) (probability of A given B)

(i) To check if A and B are mutually exclusive:
Two events are mutually exclusive if \( P(A \cap B) = 0 \). If they are mutually exclusive, they cannot happen at the same time.
We know the formula for conditional probability: \( P(B/A) = \frac{P(A \cap B)}{P(A)} \).
We can rearrange this to find \( P(A \cap B) \):
\( P(A \cap B) = P(B/A) \times P(A) \).
Substitute the given values:
\( P(A \cap B) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \).

Since \( P(A \cap B) = \frac{1}{8} \), which is not equal to 0, events A and B are not mutually exclusive.

(ii) To check if A and B are independent:
Two events are independent if \( P(A \cap B) = P(A) \times P(B) \). Also, for independent events, \( P(B/A) = P(B) \) and \( P(A/B) = P(A) \).

Let's use the second condition. From the given values:
\( P(A/B) = \frac{1}{4} \).
We are also given \( P(A) = \frac{1}{4} \).
Since \( P(A/B) = P(A) \), this condition for independence is satisfied. This is a quick way to check.

Alternatively, we can calculate \( P(B) \) using \( P(A/B) = \frac{P(A \cap B)}{P(B)} \).
We know \( P(A \cap B) = \frac{1}{8} \) and \( P(A/B) = \frac{1}{4} \).
\( \frac{1}{4} = \frac{1/8}{P(B)} \)
\( P(B) = \frac{1/8}{1/4} = \frac{1}{8} \times \frac{4}{1} = \frac{4}{8} = \frac{1}{2} \).

Now, let's check the main condition for independence: \( P(A \cap B) = P(A) \times P(B) \).
\( P(A) \times P(B) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \).
Since \( P(A \cap B) = \frac{1}{8} \) and \( P(A) \times P(B) = \frac{1}{8} \), the condition is satisfied.
Therefore, events A and B are independent events.
In simple words: For part (i), we found the chance of both A and B happening together. Since this chance was not zero, they are not mutually exclusive (they can happen at the same time). For part (ii), we checked if the chance of A happening given B had already happened was the same as the chance of A happening by itself. Since it was, it means A and B don't affect each other, so they are independent.

🎯 Exam Tip: Distinguish carefully between "mutually exclusive" (\( P(A \cap B) = 0 \)) and "independent" (\( P(A \cap B) = P(A)P(B) \)). Often, if events are mutually exclusive and have non-zero probabilities, they cannot be independent. Also, using \( P(A/B) = P(A) \) (or \( P(B/A) = P(B) \)) is a quick test for independence.

Question 11. (i) An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shot are 0.4,0.3,0.2 and 0.1 respectively. What is the probability that the gun hits the plane?
(ii) The probability that an event A happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. Find the probability that the event A happens at least once.
(iii) A man can kill a bird once in three shots. On the assumption that he fires three shots, what is the chance that a bird is killed?
(iv) A piece of equipment will function only when all three parts A, B, C are working. The probability of part A failing during one year is \( \frac { 1 }{ 6 } \), that of B failing is \( \frac { 1 }{ 20 } \) and that of C failing is \( \frac { 1 }{ 10 } \). What is the probability that the equipment will fail before the end of the year?
(v) The probability that A hits a target is 1/3 and the probability that B hits it is 2/5. What is the probability that the target will be hit, if each one M and B shoots at the target?
Answer:
(i) Let \( E_1, E_2, E_3, E_4 \) be the events that the gun hits the plane on the first, second, third, and fourth shots, respectively. The chance of hitting changes with each shot.
Given probabilities of hitting:
\( P(E_1) = 0.4 \)
\( P(E_2) = 0.3 \)
\( P(E_3) = 0.2 \)
\( P(E_4) = 0.1 \)

The problem asks for the probability that the gun *hits* the plane at least once. It's often easier to calculate the probability that it *misses* every time and subtract that from 1.
Probability of missing on each shot:
\( P(\overline{E_1}) = 1 - P(E_1) = 1 - 0.4 = 0.6 \)
\( P(\overline{E_2}) = 1 - P(E_2) = 1 - 0.3 = 0.7 \)
\( P(\overline{E_3}) = 1 - P(E_3) = 1 - 0.2 = 0.8 \)
\( P(\overline{E_4}) = 1 - P(E_4) = 1 - 0.1 = 0.9 \)

Assuming the shots are independent, the probability that the gun misses the plane on all four shots is:
\( P(\text{misses all}) = P(\overline{E_1}) \times P(\overline{E_2}) \times P(\overline{E_3}) \times P(\overline{E_4}) \)
\( = 0.6 \times 0.7 \times 0.8 \times 0.9 \)
\( = 0.3024 \).

The required probability that the gun hits the plane (at least once) is:
\( P(\text{hits plane}) = 1 - P(\text{misses all}) \)
\( = 1 - 0.3024 = 0.6976 \).

(ii) Let E be the event that event A happens in one trial. The probability of A happening is \( P(E) = 0.4 \).
Three independent trials are performed. The trials are separate, so outcomes don't affect each other.
Probability that event A does *not* happen in one trial is \( P(\overline{E}) = 1 - P(E) = 1 - 0.4 = 0.6 \).

We need the probability that event A happens at least once in three trials.
It's easier to find the probability that A *never* happens in three trials and subtract it from 1.
Probability that A never happens in three trials \( = P(\overline{E}) \times P(\overline{E}) \times P(\overline{E}) \)
\( = (0.6)^3 = 0.216 \).

Required probability (A happens at least once) \( = 1 - P(\text{A never happens}) \)
\( = 1 - 0.216 = 0.784 \).

(iii) A man can kill a bird once in three shots. This means the probability of success (killing the bird in one shot) is \( p = \frac{1}{3} \).
The probability of failure (not killing the bird in one shot) is \( q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \).
He fires three shots. We want the chance that the bird is killed (at least once). This is a common application of probability in real-world scenarios.

Again, it's easier to find the probability that the bird is *not* killed in any of the three shots and subtract it from 1.
Probability that the bird is not killed in one shot is \( q = \frac{2}{3} \).
Probability that the bird is not killed in all three shots (qqq), assuming shots are independent:
\( P(\text{not killed in 3 shots}) = q \times q \times q = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \).

The required probability that the bird is killed (at least once) is:
\( P(\text{bird killed}) = 1 - P(\text{not killed in 3 shots}) \)
\( = 1 - \frac{8}{27} = \frac{19}{27} \).

(iv) An equipment functions only when all three parts A, B, C are working.
Given probabilities of failure for each part:
\( P(\text{A fails}) = P(\overline{A}) = \frac{1}{6} \)
\( P(\text{B fails}) = P(\overline{B}) = \frac{1}{20} \)
\( P(\text{C fails}) = P(\overline{C}) = \frac{1}{10} \)

Probabilities of each part working:
\( P(A) = 1 - P(\overline{A}) = 1 - \frac{1}{6} = \frac{5}{6} \)
\( P(B) = 1 - P(\overline{B}) = 1 - \frac{1}{20} = \frac{19}{20} \)
\( P(C) = 1 - P(\overline{C}) = 1 - \frac{1}{10} = \frac{9}{10} \)

The equipment works only if all three parts A, B, and C are working. Assuming the parts fail independently, the probability that the equipment *works* is:
\( P(\text{equipment works}) = P(A) \times P(B) \times P(C) \)
\( = \frac{5}{6} \times \frac{19}{20} \times \frac{9}{10} \)
\( = \frac{5 \times 19 \times 9}{6 \times 20 \times 10} = \frac{5 \times 19 \times 9}{1200} = \frac{855}{1200} \)
\( = \frac{171}{240} = \frac{57}{80} \).

We need the probability that the equipment *fails* before the end of the year.
\( P(\text{equipment fails}) = 1 - P(\text{equipment works}) \)
\( = 1 - \frac{57}{80} = \frac{80 - 57}{80} = \frac{23}{80} \).

(v) Probability that A hits a target is \( P(A_{\text{hit}}) = \frac{1}{3} \).
Probability that A misses the target is \( P(A_{\text{miss}}) = 1 - \frac{1}{3} = \frac{2}{3} \).

Probability that B hits a target is \( P(B_{\text{hit}}) = \frac{2}{5} \).
Probability that B misses the target is \( P(B_{\text{miss}}) = 1 - \frac{2}{5} = \frac{3}{5} \).

We want the probability that the target will be hit, if each one (A and B) shoots at the target. This means the target is hit if A hits, or B hits, or both hit. This is another situation where calculating the complement is simpler.
The target is *not* hit if both A and B miss the target. The events are independent.
\( P(\text{target not hit}) = P(A_{\text{miss}}) \times P(B_{\text{miss}}) \)
\( = \frac{2}{3} \times \frac{3}{5} = \frac{6}{15} = \frac{2}{5} \).

The required probability that the target will be hit is:
\( P(\text{target hit}) = 1 - P(\text{target not hit}) \)
\( = 1 - \frac{2}{5} = \frac{3}{5} \).
In simple words: For (i), (ii), (iii), and (v), it was easier to find the chance that an event *doesn't* happen (like missing the plane or not killing the bird) and then subtract that from 1 to get the chance that it happens *at least once*. For (iv), to find the chance the equipment fails, we first found the chance that all parts work together and then subtracted that from 1. These problems often rely on the concept of complementary probability.

🎯 Exam Tip: For "at least once" or "at least one" problems, it is usually more efficient to calculate the probability of the complementary event (i.e., "never" or "none") and subtract it from 1. Ensure all events are correctly identified as independent or dependent.

Question 12. (i) Probability that man will be alive 25 years hence is 0.3 and the probability that his wife will be alive 25 years hence is 0.4. Find the probability that 25 years hence (a) both will be alive (b) only the man will be alive (c) only the woman will be alive (d) at least one of them will be alive.
(ii) The probability that Krishna will be alive 10 years hence is \( \frac { 7 }{ 15 } \) and Hari will be alive is \( \frac { 7 }{ 10 } \). Find the probability that both Krishna and Hari will be dead 10 years hence.
Answer:
(i) Let A be the event that the man will be alive 25 years hence.
Let B be the event that his wife will be alive 25 years hence.
Given: \( P(A) = 0.3 \)
Given: \( P(B) = 0.4 \)
We assume the events of the man and his wife being alive are independent. This is a reasonable assumption in probability problems unless stated otherwise.

Probability that the man will *not* be alive (dead) = \( P(\overline{A}) = 1 - P(A) = 1 - 0.3 = 0.7 \).
Probability that the wife will *not* be alive (dead) = \( P(\overline{B}) = 1 - P(B) = 1 - 0.4 = 0.6 \).

(a) Probability that both will be alive:
Since A and B are independent, \( P(A \cap B) = P(A) \times P(B) \).
\( = 0.3 \times 0.4 = 0.12 \).

(b) Probability that only the man will be alive:
This means the man is alive (A) AND the wife is not alive (\( \overline{B} \)).
\( P(A \cap \overline{B}) = P(A) \times P(\overline{B}) \) (due to independence).
\( = 0.3 \times 0.6 = 0.18 \).

(c) Probability that only the woman will be alive:
This means the woman is alive (B) AND the man is not alive (\( \overline{A} \)).
\( P(\overline{A} \cap B) = P(\overline{A}) \times P(B) \) (due to independence).
\( = 0.7 \times 0.4 = 0.28 \).

(d) Probability that at least one of them will be alive:
This means the man is alive, or the wife is alive, or both are alive. It's easier to find the complement: neither of them is alive.
\( P(\text{at least one alive}) = 1 - P(\text{neither alive}) \)
\( P(\text{neither alive}) = P(\overline{A} \cap \overline{B}) = P(\overline{A}) \times P(\overline{B}) \) (due to independence).
\( = 0.7 \times 0.6 = 0.42 \).
So, \( P(\text{at least one alive}) = 1 - 0.42 = 0.58 \).

(ii) Let K be the event that Krishna will be alive 10 years hence.
Let H be the event that Hari will be alive 10 years hence.
Given: \( P(K) = \frac{7}{15} \).
Given: \( P(H) = \frac{7}{10} \).
We need to find the probability that both Krishna and Hari will be dead 10 years hence. This means neither K nor H happens.

Probability that Krishna will be dead 10 years hence:
\( P(\overline{K}) = 1 - P(K) = 1 - \frac{7}{15} = \frac{15 - 7}{15} = \frac{8}{15} \).

Probability that Hari will be dead 10 years hence:
\( P(\overline{H}) = 1 - P(H) = 1 - \frac{7}{10} = \frac{10 - 7}{10} = \frac{3}{10} \).

Assuming the events of Krishna and Hari being alive (or dead) are independent, which is standard for such problems.
The required probability that both Krishna and Hari will be dead 10 years hence is:
\( P(\overline{K} \cap \overline{H}) = P(\overline{K}) \times P(\overline{H}) \)
\( = \frac{8}{15} \times \frac{3}{10} \)
\( = \frac{24}{150} \).
Simplify the fraction:
\( = \frac{4}{25} \).
In simple words: For part (i), we used the individual chances of the man and wife being alive or dead, assuming their fates are separate. We then multiplied these chances to find the probability of different combined outcomes, like both being alive or only one. For part (ii), we first found the chances of Krishna and Hari being dead, and then multiplied them to find the chance of both being dead, assuming their lives are independent.

🎯 Exam Tip: Always clearly define events and their complements. For problems involving "only A" or "at least one," draw a small Venn diagram or use the complement rule to correctly combine probabilities of independent events.

Question 13. A bag contains 3 white, 2 black and 4 red balls. Find the probability of drawing a white, a black and a red ball in succession in that order.
Answer: Given the number of balls in the bag:
White balls = 3
Black balls = 2
Red balls = 4
Total number of balls in the bag is \( 3 + 2 + 4 = 9 \). This mix of balls determines the probabilities.

This question has two possible interpretations: with replacement or without replacement. Standard practice is to assume "without replacement" unless "with replacement" is explicitly stated, but the solution provides both. Let's cover both scenarios.

**Scenario 1: When balls are drawn with replacement.**
This means after each ball is drawn, it is put back into the bag. So, the total number of balls and the number of each color remain constant for every draw.

Probability of drawing a white ball first: \( P(W_1) = \frac{3}{9} \).
Probability of drawing a black ball second: \( P(B_2) = \frac{2}{9} \).
Probability of drawing a red ball third: \( P(R_3) = \frac{4}{9} \).

The required probability of drawing a white, a black, and a red ball in succession (with replacement) is:
\( = P(W_1) \times P(B_2) \times P(R_3) \)
\( = \frac{3}{9} \times \frac{2}{9} \times \frac{4}{9} \)
\( = \frac{24}{729} = \frac{8}{243} \).

**Scenario 2: When balls are drawn without replacement.**
This means after each ball is drawn, it is *not* put back. So, the total number of balls decreases, and the number of balls of a specific color also decreases if that color was drawn.

Probability of drawing a white ball first:
\( P(W_1) = \frac{3}{9} \).
After drawing one white ball, there are 8 balls left (2 white, 2 black, 4 red).

Probability of drawing a black ball second (given the first was white and not replaced):
\( P(B_2|W_1) = \frac{2}{8} \).
After drawing one white and one black ball, there are 7 balls left (2 white, 1 black, 4 red).

Probability of drawing a red ball third (given the first two were white and black, and not replaced):
\( P(R_3|W_1 \cap B_2) = \frac{4}{7} \).

The required probability of drawing a white, a black, and a red ball in succession (without replacement) is:
\( = P(W_1) \times P(B_2|W_1) \times P(R_3|W_1 \cap B_2) \)
\( = \frac{3}{9} \times \frac{2}{8} \times \frac{4}{7} \).
Simplify the fractions:
\( = \frac{1}{3} \times \frac{1}{4} \times \frac{4}{7} \)
\( = \frac{1 \times 1 \times 4}{3 \times 4 \times 7} = \frac{4}{84} \)
\( = \frac{1}{21} \).
In simple words: This problem can be solved in two ways. If you put the ball back after each pick, the chances stay the same for each color, so you just multiply the starting chances. If you don't put the ball back, the number of balls and the number of colors available decrease each time, making the chances change. In that case, you multiply the changing chances.

🎯 Exam Tip: Always clarify whether "with replacement" or "without replacement" is implied. This distinction fundamentally changes the probabilities for successive draws, as it determines whether events are independent or dependent.

Question 14. (i) A problem in physics is given to three students A, B, C whose chances of Solving it are \( \frac { 1 }{ 3 } \), \( \frac { 1 }{ 4 } \), \( \frac { 1 }{ 2 } \) respectively. Find the probability that the problem will be Solved.
(ii) Three persons A, B, C can Solve a problem independently with probabilities \( \frac { 1 }{ 3 } \), \( \frac { 1 }{ 4 } \) and \( \frac { 1 }{ 5 } \). Find the probability that problem is Solved.
Answer:
(i) Let A, B, and C be the events that students A, B, and C solve the problem, respectively. The chances of solving a problem vary among students.
Given probabilities of solving the problem:
\( P(A) = \frac{1}{3} \)
\( P(B) = \frac{1}{4} \)
\( P(C) = \frac{1}{2} \)

We need to find the probability that the problem will be solved. This means at least one of the students solves it.
It's easier to find the probability that *none* of them solve the problem and subtract it from 1.

Probabilities of *not* solving the problem:
\( P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3} \)
\( P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4} \)
\( P(\overline{C}) = 1 - P(C) = 1 - \frac{1}{2} = \frac{1}{2} \)

Assuming they work independently (as is common in such problems), the probability that *none* of them solve the problem is:
\( P(\overline{A} \cap \overline{B} \cap \overline{C}) = P(\overline{A}) \times P(\overline{B}) \times P(\overline{C}) \)
\( = \frac{2}{3} \times \frac{3}{4} \times \frac{1}{2} \)
\( = \frac{6}{24} = \frac{1}{4} \).

The required probability that the problem will be solved (at least one solves it) is:
\( P(\text{problem solved}) = 1 - P(\text{none solve it}) \)
\( = 1 - \frac{1}{4} = \frac{3}{4} \).

(ii) Let A, B, and C be the events that persons A, B, and C solve the problem, respectively.
Given probabilities of solving the problem independently:
\( P(A) = \frac{1}{3} \)
\( P(B) = \frac{1}{4} \)
\( P(C) = \frac{1}{5} \)

Again, we need the probability that the problem is solved (at least one person solves it). We use the complement approach.

Probabilities of *not* solving the problem:
\( P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3} \)
\( P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4} \)
\( P(\overline{C}) = 1 - P(C) = 1 - \frac{1}{5} = \frac{4}{5} \)

Since they solve the problem independently, the probability that *none* of them solve the problem is:
\( P(\overline{A} \cap \overline{B} \cap \overline{C}) = P(\overline{A}) \times P(\overline{B}) \times P(\overline{C}) \)
\( = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \).
\( = \frac{2 \times 3 \times 4}{3 \times 4 \times 5} = \frac{24}{60} = \frac{2}{5} \).

The required probability that the problem will be solved (at least one solves it) is:
\( P(\text{problem solved}) = 1 - P(\text{none solve it}) \)
\( = 1 - \frac{2}{5} = \frac{3}{5} \).
In simple words: For both parts, when you want to know the chance that a task gets done by at least one person, it's simpler to first find the chance that *nobody* does it. Then, you subtract that chance from 1 to get your answer. This works because the students/persons work on the problem separately without influencing each other.

🎯 Exam Tip: When multiple independent events lead to a successful outcome if 'at least one' occurs, the most efficient method is to calculate the probability of the complementary event (none of them succeed) and subtract it from 1. This significantly reduces computation compared to summing individual success cases.