GSEB Class 12 Maths Solutions Chapter 13 Probability Exercise 13.2

Question 1. If \( P(A) = \frac{3}{5} \) and \( P(B) = \frac{1}{5} \), find \( P(A \cap B) \), if A and B are independent events.
Answer: Because events A and B are independent, we find the probability of their intersection. This simply means we multiply their individual probabilities.
\( P(A \cap B) = P(A) \times P(B) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{25}. \)
In simple words: When two events don't affect each other, you can find the chance of both happening by multiplying their separate chances.

Exam Tip: Remember the fundamental definition for independent events: \( P(A \cap B) = P(A) \times P(B) \). Always apply this formula directly when independence is stated.

Question 2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Answer:
Number of total possible outcomes = 52
Number of black cards available = 26
A single black card can be selected in 26 different ways.
Probability of obtaining the first black card \( P(A) = \frac{26}{52} = \frac{1}{2} \).
After selecting one card, the remaining number of cards is 51.
After drawing one black card, the remaining number of black cards is 25.
Therefore, the probability of getting both black cards is calculated as:
\( P(A) P(B/A) = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}. \)
In simple words: First, find the chance of drawing a black card. Then, since you didn't put the first card back, there are fewer cards left. So, find the chance of drawing another black card from the remaining pile and multiply the two chances together.

Exam Tip: For problems "without replacement," ensure you adjust the total number of items and the number of favorable items for each subsequent draw. This is crucial for accurate probability calculation.

Question 3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise it is rejected. Find the probability that a box containing 15 oranges, out of which 12 are good and three are bad ones, will be approved for sale.
Answer: The box contains 12 good oranges and 3 bad oranges. The box gets approved for sale if all three selected oranges are good ones.
The number of ways to choose 3 good oranges is \( ^{12}C_3 \).
The total number of ways to select 3 oranges out of 15 is \( ^{15}C_3 \).
The probability that the box is approved is the probability of selecting 3 good oranges.
\( = \frac{{}^{12} C_{3}}{{}^{15} C_{3}} = \frac{12 \times 11 \times 10}{15 \times 14 \times 13} = \frac{44}{91}. \)
In simple words: We need to pick three good oranges out of the good ones available, and divide that by the total ways to pick any three oranges from the box. This gives us the chance the box will be approved.

Exam Tip: When selecting multiple items "without replacement" and the order doesn't matter, use combinations (nCr). The probability is the ratio of favorable combinations to total combinations.

Question 4. A fair coin and an unbiased dice are tossed. Let A be the event 'head appears on the coin' and B be the event 3 on dice. Check whether A and B are independent events or not.
Answer: When a coin is flipped, either a head or a tail will appear.
The probability of obtaining a head \( P(A) = \frac{1}{2} \).
When a dice is rolled, any number from 1, 2, 3, 4, 5, 6 will show up.
The probability of getting a 3, \( P(B) = \frac{1}{6} \).
When both a coin and a dice are tossed, the total possible outcomes are:
H1, H2, H3, H4, H5, H6
T1, T2, T3, T4, T5, T6
The event 'head and 3' will happen in only 1 way (H3).
The probability of getting 'head and 3' \( = \frac{1}{12} \).
This means \( P(A \cap B) = \frac{1}{12} \).
Now, let's calculate \( P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \).
Since \( P(A \cap B) = P(A) \times P(B) \), it implies that events A and B are independent.
In simple words: The chance of getting a head on a coin is half, and the chance of getting a 3 on a dice is one-sixth. If you multiply these chances, you get the chance of both happening together. Since this matches the actual chance of both happening, the events are independent.

Exam Tip: To prove independence, always calculate \( P(A \cap B) \) and \( P(A) \times P(B) \) separately. If they are equal, the events are independent; if not, they are dependent.

Question 5. A dice, marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event 'number is even' and B be the event 'number is red'. Are A and B are independent?
Answer: The even numbers on the dice are 2, 4, 6.
The probability of getting an even number \( P(A) = \frac{3}{6} = \frac{1}{2} \).
There are two colors on the dice: red and green.
The probability of getting a red color number \( P(B) = \frac{3}{6} = \frac{1}{2} \).
The even number that is also red is 2.
Therefore, the probability of getting a red color and an even number \( P(A \cap B) = \frac{1}{6} \).
Let's calculate the product \( P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
Since \( \frac{1}{4} \ne \frac{1}{6} \), it means \( P(A) \times P(B) \ne P(A \cap B) \).
This implies that A and B are not independent events.
In simple words: We check if the chance of getting an even number and a red number is the same as multiplying the chance of getting an even number by the chance of getting a red number. Since they are not the same, these two events affect each other and are not independent.

Exam Tip: Carefully list all possible outcomes for each event and their intersection. Always compare \( P(A \cap B) \) directly with \( P(A) \times P(B) \) to determine independence.

Question 6. Let E and F be the events with \( P(E) = \frac{3}{5}, P(F) = \frac{3}{10} \) and \( P(E \cap F) = \frac{1}{5} \). Are E and F independent?
Answer: We are given the following probabilities:
\( P(E) = \frac{3}{5} \)
\( P(F) = \frac{3}{10} \)
Let's calculate the product of their individual probabilities:
\( P(E) \times P(F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50} \).
We are also given \( P(E \cap F) = \frac{1}{5} \).
We can see that \( P(E \cap F) \ne P(E) \times P(F) \) since \( \frac{1}{5} \ne \frac{9}{50} \).
Therefore, the events E and F are not independent.
In simple words: To see if two events are independent, we multiply their individual chances. If this result is different from the chance of both events happening together, then they are not independent.

Exam Tip: The crucial step is comparing the calculated product \( P(E) \times P(F) \) with the given \( P(E \cap F) \). A mismatch immediately indicates dependence.

Question 7. Given that the events A and B are such that \( P(A) = \frac{1}{2}, P(A \cup B) = \frac{3}{5} \) and \( P(B) = p \), If they are
(i) mutually exclusive
(ii) independent.
Answer: We begin by using the general formula for the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute the given values into the formula:
\( \frac{3}{5} = \frac{1}{2} + p - x \), where we let \( P(A \cap B) = x \).
Rearranging the equation to solve for \( p - x \):
\( p - x = \frac{3}{5} - \frac{1}{2} \)
\( p - x = \frac{6-5}{10} \)
\( p - x = \frac{1}{10} \) ...(1)

(i) When events A and B are mutually exclusive, their intersection is empty, meaning \( x = P(A \cap B) = 0 \).
Substitute \( x = 0 \) into equation (1):
\( p - 0 = \frac{1}{10} \)
\( p = \frac{1}{10} \).
(ii) When events A and B are independent, their intersection is the product of their individual probabilities:
\( P(A \cap B) = P(A) \times P(B) \)
So, \( x = \frac{1}{2} \times p \). ...(2)
Now, substitute this expression for \( x \) into equation (1):
\( p - \frac{1}{2}p = \frac{1}{10} \)
\( \frac{1}{2}p = \frac{1}{10} \)
To find \( p \), multiply both sides by 2:
\( p = \frac{2}{10} \)
\( p = \frac{1}{5} \).
In simple words: First, we use the rule for the probability of A or B happening. If they are mutually exclusive, they can't happen at the same time, so their shared probability is zero. If they are independent, the chance of both happening is just their chances multiplied together. We use these ideas to find the value of 'p'.

Exam Tip: Clearly differentiate between mutually exclusive (no common outcomes, \( P(A \cap B) = 0 \)) and independent events (occurrence of one doesn't affect the other, \( P(A \cap B) = P(A) \times P(B) \)). This distinction is key to solving such problems correctly.

Question 8. Let A and B be independent events \( P(A) = 0.3 \) and \( P(B) = 0.4 \). Find:
(i) \( P(A \cap B) \)
(ii) \( P(A \cup B) \)
(iii) \( P(A/B) \)
(iv) \( P(B/A) \)
Answer: We are given \( P(A) = 0.3 \) and \( P(B) = 0.4 \). Since A and B are independent events, we can find the required probabilities.
(i) For independent events, the probability of their intersection is the product of their individual probabilities:
\( P(A \cap B) = P(A) \times P(B) \)
\( = 0.3 \times 0.4 = 0.12 \).
(ii) The probability of the union of two events is given by:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Using the result from part (i):
\( = 0.3 + 0.4 - 0.12 = 0.7 - 0.12 = 0.58 \).
(iii) The conditional probability \( P(A/B) \) for independent events is simply \( P(A) \). However, we can also calculate it using the formula:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
Using the result from part (i):
\( = \frac{0.12}{0.4} = \frac{12}{40} = \frac{3}{10} = 0.3 \).
(iv) Similarly, the conditional probability \( P(B/A) \) for independent events is simply \( P(B) \). Using the formula:
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
Using the result from part (i):
\( = \frac{0.12}{0.3} = \frac{12}{30} = \frac{2}{5} = 0.4 \).
In simple words: When two events are independent, the chance of both happening is their chances multiplied. The chance of either happening is their sum minus the chance of both. And the chance of one happening given the other already happened is just the chance of the first event itself.

Exam Tip: For independent events, remember these key shortcuts: \( P(A \cap B) = P(A) \times P(B) \), \( P(A/B) = P(A) \), and \( P(B/A) = P(B) \). Use the general union formula for \( P(A \cup B) \).

Question 9. If A and B are two events, such that \( P(A) = \frac{1}{4}, P(B) = \frac{1}{2}, P(A \cap B) = \frac{1}{8} \), find P(not A and not B).
Answer: We want to find the probability of 'not A and not B', which can be written as \( P(A' \cap B') \). By De Morgan's laws, this is equivalent to \( P((A \cup B)') \).
The probability of the complement of an event is 1 minus the probability of the event itself:
\( P((A \cup B)') = 1 - P(A \cup B) \).
First, let's calculate \( P(A \cup B) \) using the formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute the given values:
\( P(A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} \)
To combine these fractions, find a common denominator, which is 8:
\( P(A \cup B) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} \)
\( P(A \cup B) = \frac{2+4-1}{8} = \frac{5}{8} \).
Now, we can find \( P(A' \cap B') \):
\( P(A' \cap B') = 1 - P(A \cup B) = 1 - \frac{5}{8} = \frac{3}{8} \).
In simple words: We want the chance that neither A nor B happens. This is the same as 1 minus the chance that A or B (or both) happen. So, we first find the chance of A or B happening using a standard formula, and then subtract that from 1.

Exam Tip: Remember De Morgan's laws: \( (A \cup B)' = A' \cap B' \) and \( (A \cap B)' = A' \cup B' \). These are very useful for simplifying complex probability expressions involving complements.

Question 10. Events A and B are such that \( P(A) = \frac{1}{2}, P(B) = \frac{7}{12} \) and P(not A or not B) = \( \frac{1}{4} \). State whether A and B are independent.
Answer: We are given \( P(A) = \frac{1}{2} \) and \( P(B) = \frac{7}{12} \).
We are also given \( P(\text{not A or not B}) = \frac{1}{4} \). Using De Morgan's laws, P(not A or not B) is \( P(A' \cup B') \), which is equivalent to \( P((A \cap B)') \).
So, \( P((A \cap B)') = \frac{1}{4} \).
We know that \( P(E') = 1 - P(E) \). Therefore:
\( P(A \cap B) = 1 - P((A \cap B)') = 1 - \frac{1}{4} = \frac{3}{4} \).
Now, let's check for independence by comparing \( P(A \cap B) \) with \( P(A) \times P(B) \).
\( P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24} \).
Since \( P(A \cap B) = \frac{3}{4} \) and \( P(A) \times P(B) = \frac{7}{24} \), we observe that \( \frac{3}{4} \ne \frac{7}{24} \).
This means \( P(A \cap B) \ne P(A) \times P(B) \).
Therefore, A and B are not independent events.
In simple words: First, we use the given information to find the chance of both A and B happening. Then, we multiply the individual chances of A and B. If these two results are different, then A and B are not independent.

Exam Tip: When given "not A or not B", immediately recognize it as \( P(A' \cup B') \) and use De Morgan's law to convert it to \( P((A \cap B)') \). This allows you to find \( P(A \cap B) \) and test for independence.

Question 11. Given two independent events A and B such that \( P(A) = 0.3 \) and \( P(B) = 0.6 \), find
(i) P(A and B)
(ii) P(A and not B)
(iii) P(A or B)
(iv) P(neither A nor B)
Answer: We are given that A and B are independent events, with \( P(A) = 0.3 \) and \( P(B) = 0.6 \).
(i) For independent events, the probability of both A and B occurring (P(A and B)) is the product of their individual probabilities:
\( P(A \cap B) = P(A) \times P(B) = 0.3 \times 0.6 = 0.18 \).
(ii) The probability of A occurring and B not occurring (P(A and not B)) is \( P(A \cap B') \). Since A and B are independent, A and B' are also independent.
\( P(A \cap B') = P(A) \times P(B') \).
First, find \( P(B') = 1 - P(B) = 1 - 0.6 = 0.4 \).
So, \( P(A \cap B') = 0.3 \times 0.4 = 0.12 \).
Alternatively, for independent events, \( P(A \cap B') = P(A) - P(A \cap B) \).
\( = 0.3 - 0.18 = 0.12 \).
(iii) The probability of A or B occurring (P(A or B)) is \( P(A \cup B) \). We use the general formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
Using the value of \( P(A \cap B) \) from part (i):
\( = 0.3 + 0.6 - 0.18 = 0.9 - 0.18 = 0.72 \).
(iv) The probability of neither A nor B occurring (P(neither A nor B)) is \( P(A' \cap B') \). Since A and B are independent, A' and B' are also independent.
So, \( P(A' \cap B') = P(A') \times P(B') \).
First, find \( P(A') = 1 - P(A) = 1 - 0.3 = 0.7 \).
And \( P(B') = 1 - P(B) = 1 - 0.6 = 0.4 \).
Thus, \( P(A' \cap B') = 0.7 \times 0.4 = 0.28 \).
Alternatively, using De Morgan's laws: \( P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \).
Using the value of \( P(A \cup B) \) from part (iii):
\( = 1 - 0.72 = 0.28 \).
In simple words: When events don't affect each other, we can find probabilities for various combinations. The chance of both happening is their chances multiplied. The chance of one happening but not the other is the chance of the first times the chance of the second not happening. The chance of either happening is the sum of their chances minus the chance of both. And the chance of neither happening is the chance of the first not happening times the chance of the second not happening.

Exam Tip: Remember that if events A and B are independent, then A and B', A' and B, and A' and B' are also independent. This simplifies calculations involving complements in independence problems.

Question 12. A dice is tossed thrice. Find the probability of getting an odd number at least once?
Answer: If a dice is rolled, an odd number can be 1, 3, or 5.
The probability of obtaining an odd number in one roll of a dice \( = \frac{3}{6} = \frac{1}{2} \).
The probability of not getting an odd number (i.e., getting an even number) in one roll \( = 1 - \frac{1}{2} = \frac{1}{2} \).
Now, we want to find the probability of getting an odd number at least once when the dice is tossed thrice. It's easier to find the complement: the probability of getting *no* odd number in three tosses.
Probability of getting no odd number in three tosses (meaning all three are even) \( = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \).
Therefore, the probability of getting at least one odd number in three tosses \( = 1 - (\text{probability of no odd number}) = 1 - \frac{1}{8} = \frac{7}{8} \).
In simple words: The chance of rolling an odd number is half. We want to know the chance of getting an odd number at least once in three rolls. It's simpler to calculate the chance of *never* getting an odd number (meaning all rolls are even) and then subtract that from 1.

Exam Tip: For "at least one" probability questions, it's often simpler to calculate the probability of the complementary event (i.e., "none") and subtract it from 1. This avoids summing multiple cases.

Question 13. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.
Answer: The box contains 10 black (B) balls and 8 red (R) balls. The total number of balls is \( 10+8=18 \). Since balls are drawn "with replacement", the probability for each draw remains constant.
(i) Probability of getting the first ball red:
\( P(\text{1st Red}) = \frac{8}{18} = \frac{4}{9} \).
Since the first ball is replaced, the probability of getting the second ball red is also \( \frac{4}{9} \).
Therefore, the probability of getting both balls red:
\( P(\text{both Red}) = P(\text{1st Red}) \times P(\text{2nd Red}) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81} \).
(ii) Probability of getting the first ball black:
\( P(\text{1st Black}) = \frac{10}{18} = \frac{5}{9} \).
Probability of getting the second ball red (with replacement):
\( P(\text{2nd Red}) = \frac{8}{18} = \frac{4}{9} \).
Therefore, the probability of getting the first ball black and the second ball red:
\( P(\text{1st Black and 2nd Red}) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81} \).
(iii) One ball is black and the other is red. This can happen in two ways: (Black then Red) OR (Red then Black).
Probability of (Black then Red) = \( P(\text{1st Black}) \times P(\text{2nd Red}) = \frac{10}{18} \times \frac{8}{18} = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81} \).
Probability of (Red then Black) = \( P(\text{1st Red}) \times P(\text{2nd Black}) = \frac{8}{18} \times \frac{10}{18} = \frac{4}{9} \times \frac{5}{9} = \frac{20}{81} \).
The total probability of getting one black and one red (in any order) is the sum of these two probabilities:
\( P(\text{one Black, one Red}) = \frac{20}{81} + \frac{20}{81} = \frac{40}{81} \).
In simple words: Since we put the ball back each time, the chances stay the same for every draw. For both red, we multiply the chance of getting red twice. For black then red, we multiply those specific chances. For one of each color, we add the chances of black-then-red and red-then-black.

Exam Tip: "With replacement" means events are independent, so probabilities don't change for subsequent draws. "Without replacement" means events are dependent, requiring adjustments to total outcomes and favorable outcomes for each draw.

Question 14. Problem of solving specific problem independently by A and B are \( \frac{1}{2} \) and \( \frac{1}{3} \) respectively. If both try to solve the problem independently, find the probability that:
(i) The problem is solved.
(ii) Exactly one of them solves the problem.
Answer: Let P(A) be the probability that A solves the problem, and P(B) be the probability that B solves the problem.
Given: \( P(A) = \frac{1}{2} \) and \( P(B) = \frac{1}{3} \).
Since A and B try to solve the problem independently, we can find the probabilities of them not solving it.
Probability that A does not solve the problem \( P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \).
Probability that B does not solve the problem \( P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \).
(i) The problem is solved if A solves it, or B solves it, or both solve it. This is \( P(A \cup B) \). It's easier to find the complement: the probability that the problem is *not* solved.
The problem is not solved if neither A nor B solves it, which is \( P(A' \cap B') \). Since A and B are independent, A' and B' are also independent.
\( P(\text{problem not solved}) = P(A' \cap B') = P(A') \times P(B') = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \).
Therefore, the probability that the problem is solved \( = 1 - P(\text{problem not solved}) = 1 - \frac{1}{3} = \frac{2}{3} \).
(ii) Exactly one of them solves the problem means (A solves and B does not) OR (B solves and A does not). This can be written as \( P(A \cap B') + P(A' \cap B) \).
Since A and B are independent, then \( A \) and \( B' \) are independent, and \( A' \) and \( B \) are independent.
\( P(A \cap B') = P(A) \times P(B') = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3} \).
\( P(A' \cap B) = P(A') \times P(B) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \).
The probability that exactly one of them solves the problem is the sum of these two probabilities:
\( = \frac{1}{3} + \frac{1}{6} \)
To add these, find a common denominator (6):
\( = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \).
In simple words: For the problem to be solved, at least one person must solve it. This is 1 minus the chance that no one solves it. For exactly one person to solve it, either A solves and B doesn't, or B solves and A doesn't. We calculate these separate chances and then add them up.

Exam Tip: For "problem solved" scenarios with independent attempts, it's often easier to calculate the probability that the problem is *not* solved (neither succeeds) and subtract from 1. For "exactly one," calculate the two mutually exclusive scenarios and sum their probabilities.

Question 15. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
(i) E: the card drawn is a spade
F: the card drawn is an ace
(ii) E: the card drawn is black
F: the card drawn is a king,
(iii) E : the card drawn is a king or queen
F : the. card drawn is a queen or jack.
Answer: In a standard deck, there are 52 cards.
(i) E: The card drawn is a spade. F: The card drawn is an ace.
Number of spade cards = 13.
Probability of drawing a spade card \( P(E) = \frac{13}{52} = \frac{1}{4} \).
Number of ace cards = 4.
Probability of drawing an ace card \( P(F) = \frac{4}{52} = \frac{1}{13} \).
The event \( E \cap F \) means the card drawn is a spade and an ace (i.e., the Ace of Spades). There is only one Ace of Spades.
Probability of drawing the Ace of Spades \( P(E \cap F) = \frac{1}{52} \).
Now, let's check for independence: \( P(E) \times P(F) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52} \).
Since \( P(E \cap F) = P(E) \times P(F) \), the events E and F are independent in this case.

(ii) E: The card drawn is black. F: The card drawn is a king.
Number of black cards = 26.
Probability of drawing a black card \( P(E) = \frac{26}{52} = \frac{1}{2} \).
Number of king cards = 4.
Probability of drawing a king card \( P(F) = \frac{4}{52} = \frac{1}{13} \).
The event \( E \cap F \) means the card drawn is a black king. There are two black kings (King of Spades, King of Clubs).
Probability of drawing a black king \( P(E \cap F) = \frac{2}{52} = \frac{1}{26} \).
Now, let's check for independence: \( P(E) \times P(F) = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26} \).
Since \( P(E \cap F) = P(E) \times P(F) \), the events E and F are independent in this case.

(iii) E: The card drawn is a king or queen. F: The card drawn is a queen or jack.
Number of kings = 4. Number of queens = 4.
Number of cards that are a king or a queen = \( 4+4=8 \).
Probability of drawing a king or queen \( P(E) = \frac{8}{52} = \frac{2}{13} \).
Number of queens = 4. Number of jacks = 4.
Number of cards that are a queen or a jack = \( 4+4=8 \).
Probability of drawing a queen or jack \( P(F) = \frac{8}{52} = \frac{2}{13} \).
The event \( E \cap F \) means the card drawn is (king or queen) AND (queen or jack). This refers to the cards that are common to both sets, which are the queens. There are 4 queens.
Probability of drawing a queen \( P(E \cap F) = \frac{4}{52} = \frac{1}{13} \).
Now, let's check for independence: \( P(E) \times P(F) = \frac{2}{13} \times \frac{2}{13} = \frac{4}{169} \).
Since \( P(E \cap F) = \frac{1}{13} \) and \( P(E) \times P(F) = \frac{4}{169} \), and \( \frac{1}{13} \ne \frac{4}{169} \), the events E and F are not independent in this case.
In simple words: We check each pair of events to see if they are independent. For events to be independent, the chance of both happening must equal the chance of the first multiplied by the chance of the second. If these numbers match, the events are independent; otherwise, they are not.

Exam Tip: For card problems, always clearly identify the total number of cards and the number of favorable outcomes for each event. Remember that for independence, \( P(E \cap F) \) must exactly equal \( P(E) \times P(F) \).

Question 16. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.
(a) Find the probability that she read neither Hindi nor English newspaper.
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper
Answer: Let H be the event that a student reads the Hindi newspaper, and E be the event that a student reads the English newspaper.
Given:
\( P(H) = 60\% = 0.6 \)
\( P(E) = 40\% = 0.4 \)
\( P(H \cap E) = 20\% = 0.2 \)
(a) Probability that she reads neither Hindi nor English newspaper.
First, find the probability that a student reads at least one newspaper, \( P(H \cup E) \):
\( P(H \cup E) = P(H) + P(E) - P(H \cap E) \)
\( = 0.6 + 0.4 - 0.2 = 1.0 - 0.2 = 0.8 \).
The probability that she reads neither Hindi nor English newspaper is the complement of reading at least one newspaper:
\( P(\text{neither H nor E}) = 1 - P(H \cup E) = 1 - 0.8 = 0.2 \).
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper. This is a conditional probability, \( P(E/H) \).
\( P(E/H) = \frac{P(E \cap H)}{P(H)} \)
\( = \frac{0.2}{0.6} = \frac{2}{6} = \frac{1}{3} \).
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper. This is a conditional probability, \( P(H/E) \).
\( P(H/E) = \frac{P(H \cap E)}{P(E)} \)
\( = \frac{0.2}{0.4} = \frac{2}{4} = \frac{1}{2} \).
In simple words: We know the chances of reading Hindi, English, or both. We use formulas to find: (a) The chance a student reads no newspaper. (b) The chance a student reads English *if* we already know they read Hindi. (c) The chance a student reads Hindi *if* we already know they read English.

Exam Tip: Carefully identify whether the question asks for a union \( (A \cup B) \), intersection \( (A \cap B) \), complement \( (A') \), or conditional probability \( (A/B) \). Conditional probability \( P(A/B) \) always uses the formula \( \frac{P(A \cap B)}{P(B)} \).

Question 17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
(A) 0
(B) \( \frac{1}{3} \)
(C) \( \frac{1}{12} \)
(D) \( \frac{1}{36} \)
Answer: (D) \( \frac{1}{36} \)
In simple words: The only even prime number is 2. So, we want the chance of rolling a 2 on the first dice and a 2 on the second dice. Since these are independent events, we multiply their individual chances.

Exam Tip: Remember that the only even prime number is 2. When rolling multiple dice, each roll is an independent event, so probabilities are multiplied.

Question 18. Two events A and B are said to be independent, if
(A) A and B are mutually exclusive.
(B) \( P(A' \cap B') = [1 – P(A)] [1 – P(B)] \)
(C) \( P(A) = P(B) \)
(D) \( P(A) + P(B) = 1 \)
Answer: (B) \( P(A' \cap B') = [1 – P(A)] [1 – P(B)] \)
In simple words: Independent events mean that the chance of A not happening and B not happening is the same as multiplying the chance of A not happening by the chance of B not happening.

Exam Tip: The fundamental definition of independence is \( P(A \cap B) = P(A)P(B) \). From this, it follows that \( A' \) and \( B' \) are also independent, so \( P(A' \cap B') = P(A')P(B') = [1-P(A)][1-P(B)] \).