OP Malhotra Class 11 Maths Solutions Chapter 22 Probability Exercise 22 (A)

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(a)

 

Question 1. What do you mean by Random Experiment ? Give two illustrations. Define sample space associated with a random experiment. Give an example.
Answer: A random experiment is a test or process where you know all the possible results beforehand, but you cannot predict which specific result will happen each time you do it. For example, when you toss a coin, you know it will be either heads or tails, but you cannot say for sure if the next toss will be heads. Another example is throwing a dice; you know the outcome will be a number from 1 to 6, but you cannot predict the exact number that will appear. These experiments show uncertainty in their outcomes despite having a clear set of possibilities.
The sample space of a random experiment is the collection of all possible results that can occur. It lists every single outcome without any repeats. For instance, if you toss a coin, the sample space is {Head, Tail}, often written as {H, T}. This set includes every possible way the coin can land. If you throw a die, the sample space is {1, 2, 3, 4, 5, 6}.
In simple words: A random experiment is one where you know all possible results, but can't guess the exact one that will happen next. The sample space is a list of all these possible results.

🎯 Exam Tip: Clearly distinguish between knowing the *set* of outcomes and knowing the *individual* outcome. Always provide a specific example when defining concepts like random experiment and sample space.

 

Question 2. What is the resulting sample space if
(i) one coin is tossed ;
(ii) two coins are tossed simultaneously ;
(iii) three coins are tossed simultaneously ?
Answer:
(i) When one coin is tossed, the sample space S = {H, T}. This means the coin can land on either Heads or Tails.
(ii) When two coins are tossed simultaneously, the sample space S = {HH, HT, TH, TT}. Here, each outcome shows the result of both coins.
(iii) When three coins are tossed simultaneously, the sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. This can be easily visualized using a tree diagram as shown below:

In simple words: The sample space is a list of all possible outcomes. For one coin, it's Heads or Tails. For two coins, it's HH, HT, TH, TT. For three coins, you get all combinations like HHH, HHT, and so on, making 8 total outcomes.

🎯 Exam Tip: When listing sample spaces, ensure all possible unique outcomes are included and no outcomes are duplicated. For multiple events, a tree diagram can help organize all possibilities.

 

Question 3. Describe the sample space of this experiment:
(i) One die is rolled ;
(ii) Two dice are rolled.
Answer:
(i) When one die is rolled, the sample space S = {1, 2, 3, 4, 5, 6}. This represents all the possible numbers that can show up on the die.
(ii) When two dice are rolled, the sample space S consists of all possible pairs where each number in the pair is from 1 to 6. This forms a total of 36 unique outcomes. Each outcome shows the result of the first die and the second die.
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
In simple words: If you roll one die, the sample space is the numbers 1 to 6. If you roll two dice, the sample space lists every possible pair of numbers you can get, like (1,1), (1,2), up to (6,6).

🎯 Exam Tip: Remember that for two dice, (1,2) is a different outcome from (2,1). Always list all combinations systematically to avoid missing any.

 

Question 4. Describe the sample space :
(i) A coin is tossed twice. If the second throw results in a tail, a die is thrown.
(ii) A coin is tossed twice. If the second throw results in a head, a die is thrown, otherwise a coin is tossed.
(iii) A coin is tossed. If it results in a head, a die is thrown. If the die shows up an even number, the die is thrown again.
Answer:
(i) For the experiment where a coin is tossed twice, and if the second throw is a tail, a die is thrown:
If the first two tosses are HH, this outcome is included as is.
If the first two tosses are HT (second is Tail), a die is thrown, leading to HT1, HT2, HT3, HT4, HT5, HT6.
If the first two tosses are TH, this outcome is included as is.
If the first two tosses are TT (second is Tail), a die is thrown, leading to TT1, TT2, TT3, TT4, TT5, TT6.
The full sample space is S = {HH, HT1, HT2, HT3, HT4, HT5, HT6, TH, TT1, TT2, TT3, TT4, TT5, TT6}.
(ii) For the experiment where a coin is tossed twice; if the second throw is a head, a die is thrown, otherwise a coin is tossed:
Possible initial coin toss pairs are HH, HT, TH, TT.
If HH (second is Head), a die is thrown: HH1, HH2, HH3, HH4, HH5, HH6.
If HT (second is Tail - "otherwise"), a coin is tossed: HTH, HTT.
If TH (second is Head), a die is thrown: TH1, TH2, TH3, TH4, TH5, TH6.
If TT (second is Tail - "otherwise"), a coin is tossed: TTH, TTT.
The full sample space is S = {HH1, HH2, HH3, HH4, HH5, HH6, HTH, HTT, TH1, TH2, TH3, TH4, TH5, TH6, TTH, TTT}.
(iii) For the experiment where a coin is tossed; if it results in a head, a die is thrown; if the die shows an even number, the die is thrown again:
If the first toss is Tail: T.
If the first toss is Head, a die is thrown: H1, H2, H3, H4, H5, H6.
Now, for H-outcomes, if the die result is even (H2, H4, H6), throw the die again:
H1 (from H then 1)
H3 (from H then 3)
H5 (from H then 5)
H21, H22, H23, H24, H25, H26 (from H then 2, then die again)
H41, H42, H43, H44, H45, H46 (from H then 4, then die again)
H61, H62, H63, H64, H65, H66 (from H then 6, then die again)
The full sample space is S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}.
In simple words: These questions describe events that happen in steps, where the next step depends on the result of the previous one. To find the sample space, you need to follow all the possible paths and list every final outcome. For example, if tossing a coin twice and the second is a tail means you throw a die, you list all outcomes like HT1, HT2, and so on.

🎯 Exam Tip: For multi-stage experiments, carefully trace all conditional paths to ensure every possible final outcome is included in the sample space. A tree diagram can be very helpful for complex scenarios.

 

Question 5. A five-sided spinner is spun and a coin is tossed.
(i) Show the combined outcomes in a space diagram and in a tree diagram.
(ii) List the combined outcomes and state the number of equally likely combined outcomes.
Answer:
(i) The combined outcomes can be shown using a space diagram (table) and a tree diagram.

(ii) The combined outcomes are listed below:
{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5)}
The total number of equally likely combined outcomes is 10.
In simple words: When you spin a five-sided spinner and toss a coin, each spin (1 to 5) can happen with either a Head or a Tail. You can show all these pairs in a grid or by drawing a tree where branches split for each choice. In total, there are 10 different possible results.

🎯 Exam Tip: For experiments with two independent events, use a space diagram (table) to easily visualize all outcomes. The number of outcomes is the product of the number of outcomes for each individual event.

 

Question 6. In a bag there are three balls ; one red, one blue and one yellow. A ball is selected, the colour is recorded and the ball is replaced. A second ball is then selected and the colour is recorded.
(i) Show in a space diagram and in a tree diagram all the possible combined outcomes.
(ii) List these combined outcomes and state the number of equally likely combined outcomes.
Answer:
(i) The combined outcomes can be shown using a space diagram (table) and a tree diagram.

(ii) The combined outcomes are: {(R, R), (R, B), (R, Y), (B, R), (B, B), (B, Y), (Y, R), (Y, B), (Y, Y)}.
The total number of equally likely combined outcomes is 9. This makes sense because there are 3 choices for the first ball and 3 choices for the second ball.
In simple words: First, you pick a ball and put it back. Then, you pick another ball. A table or a branched diagram shows all 9 possible pairs, like picking Red then Red, or Blue then Yellow.

🎯 Exam Tip: When an item is replaced after being selected (sampling with replacement), the possibilities for the second selection remain the same as the first. This leads to a higher number of total outcomes.

 

Question 7. Satish and Mukesh who live in London wish to go on a holiday to France. They can travel to the coast by car, coach or train, and then cross the channel by ferry, train, helicopter or hovercraft.
(i) In a space diagram and in a tree diagram show all the combined outcomes of the different ways they could travel to France.
(ii) How many different ways could they travel ?
Answer:
(i) The combined outcomes of travel can be shown using a space diagram (table) and a tree diagram.

(ii) The number of different ways they could travel is calculated by multiplying the number of options for each stage of travel.
Number of ways = (Ways to travel to coast) x (Ways to cross channel)
Number of ways = 3 x 4 = 12.
Thus, the required total number of ways is 12.
In simple words: Satish and Mukesh have 3 choices to reach the coast (car, coach, train) and 4 choices to cross the channel (ferry, train, helicopter, hovercraft). To find all the possible travel plans, you multiply the number of choices for each part. This gives them 12 different ways to go to France.

🎯 Exam Tip: For sequential events, the total number of outcomes is the product of the number of options at each stage. This is a fundamental principle of counting in probability.

 

Question 8. From a group of 2 men and 3 women, two persons are selected. Describe the sample space of the experiment. If E is the event in which one man and one woman are selected, then which are the cases favourable to E ?
Answer:
Let the two men be \( M_1 \) and \( M_2 \).
Let the three women be \( W_1 \), \( W_2 \), and \( W_3 \).
When two persons are selected from this group of 5 (2 men + 3 women), the sample space (S) includes all possible pairs of people. These pairs are:
S = { \( M_1M_2 \), \( M_1W_1 \), \( M_1W_2 \), \( M_1W_3 \), \( M_2W_1 \), \( M_2W_2 \), \( M_2W_3 \), \( W_1W_2 \), \( W_2W_3 \), \( W_1W_3 \) }
There are a total of 10 possible pairs in the sample space.
Now, for event E, where one man and one woman are selected, we list the pairs that fit this condition. This event means we choose one person from the men's group and one person from the women's group.
E = { \( M_1W_1 \), \( M_1W_2 \), \( M_1W_3 \), \( M_2W_1 \), \( M_2W_2 \), \( M_2W_3 \) }
There are 6 cases favorable to event E. This reflects choosing one man out of two and one woman out of three.
In simple words: If you pick two people from a group of 2 men and 3 women, the sample space lists all 10 possible pairs you can get. If you want to pick exactly one man and one woman, there are 6 such pairs.

🎯 Exam Tip: When selecting a group of people, remember to list all combinations carefully. For "one man and one woman" events, consider all pairings between individuals from each group.

 

Question 9. A coin is tossed. If it results in a head, a coin is tossed, otherwise a die is thrown. Describe the following events:
(i) A: getting at least one head ;
(ii) B: getting an even number ;
(iii) C: Getting a tail;
(iv) D: getting a tail and an odd number.
Answer:
First, let's determine the sample space (S) for the experiment:
If the first toss is Head (H), then a coin is tossed again. The outcomes are HH, HT.
If the first toss is Tail (T), then a die is thrown. The outcomes are T1, T2, T3, T4, T5, T6.
So, the complete sample space S = {HH, HT, T1, T2, T3, T4, T5, T6}. This sample space has 8 possible outcomes.
Now, we describe the given events:
(i) Event A: getting at least one head.
This means any outcome that includes an 'H'. From the sample space, these are the outcomes where the first coin was a Head, or the second coin was a Head if the first was Head.
A = {HH, HT}
(ii) Event B: getting an even number.
This refers to outcomes where a die was thrown and it landed on an even number. This only happens if the first coin toss was a Tail.
B = {T2, T4, T6}
(iii) Event C: Getting a tail.
This means any outcome that includes a 'T'. This includes outcomes where the first coin was a Tail (T1, T2, etc.) or the second coin was a Tail (HT).
C = {HT, T1, T2, T3, T4, T5, T6}
(iv) Event D: getting a tail and an odd number.
This implies that the first coin toss was a Tail, and the die subsequently thrown showed an odd number.
D = {T1, T3, T5}
In simple words: For this experiment, the sample space includes outcomes like HH, HT (if the first toss was Head), and T1, T2, T3, T4, T5, T6 (if the first toss was Tail). Then, you list the outcomes that match each event: A for at least one head, B for an even number from the die, C for any tail, and D for a tail and an odd number from the die.

🎯 Exam Tip: Always define the full sample space first for multi-stage experiments. Then, identify the outcomes that satisfy the conditions of each specific event. Pay close attention to "at least one" vs. "exactly one" and "and" vs. "or" conditions.

 

Question 10. A coin and a die are tossed. Describe the following events.
(i) A: getting a head and an even number ;
(ii) B: getting a prime number ;
(iii) C: getting a tail and an odd number;
(iv) D: getting a head or a tail.
Answer:
First, let's determine the sample space (S) for tossing a coin and a die simultaneously.
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}. This sample space has 12 equally likely outcomes.
Now, we describe the given events:
(i) Event A: getting a head and an even number.
This means the coin shows 'H' and the die shows an even number (2, 4, or 6).
A = {H2, H4, H6}
(ii) Event B: getting a prime number.
A prime number on a die is 2, 3, or 5. This event means the die shows one of these prime numbers, regardless of the coin toss.
B = {H2, H3, H5, T2, T3, T5}
(iii) Event C: getting a tail and an odd number.
This means the coin shows 'T' and the die shows an odd number (1, 3, or 5).
C = {T1, T3, T5}
(iv) Event D: getting a head or a tail.
This event means the coin can be either a head or a tail. Since every outcome in the sample space involves either a head or a tail, this event includes all outcomes in the sample space.
D = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
In simple words: When a coin and a die are tossed together, the sample space has 12 possible results like H1, H2, T1, T2, etc. Event A is getting a head and an even number from the die. Event B is getting any prime number from the die. Event C is getting a tail and an odd number from the die. Event D includes all possible outcomes where you get either a head or a tail, which means it includes every outcome in the whole sample space.

🎯 Exam Tip: Be precise with the definitions of "and" (both conditions must be met) and "or" (at least one condition must be met) when identifying outcomes for an event. Remember that prime numbers on a die are 2, 3, and 5.

 

Question 11. A fair coin is tossed. If it shows a head, we draw a ball from a bag consisting of 3 distinct red and 4 distinct black balls, if it shows a tail, we throw a fair die. Draw a tree diagram to show all the possible outcomes and obtain .he sample space. What are sets representing the following events:
(i) the ball drawn is black ;
(ii) the coin shows tail.
Answer:
Here is the tree diagram illustrating all possible outcomes for this experiment:

The sample space S, consisting of all possible outcomes, is:
S = {HR1, HR2, HR3, HB1, HB2, HB3, HB4, T1, T2, T3, T4, T5, T6}.
There are 7 outcomes if the coin shows Head (3 Red balls + 4 Black balls) and 6 outcomes if the coin shows Tail (6 die faces), so a total of 13 outcomes.
Now, let's identify the sets for the given events:
(i) The ball drawn is black.
This event occurs when the coin shows a Head, and then a black ball is drawn from the bag.
Set for (i) = {HB1, HB2, HB3, HB4}
(ii) The coin shows tail.
This event occurs when the coin shows a Tail, and then a die is thrown, resulting in any number from 1 to 6.
Set for (ii) = {T1, T2, T3, T4, T5, T6}
In simple words: First, a coin is tossed. If it's heads, you pick one of 7 balls (3 red, 4 black). If it's tails, you roll a die (1-6). The tree diagram shows all these choices. The sample space is a list of every final outcome. Then, you list outcomes where a black ball was drawn, and outcomes where the coin showed a tail.

🎯 Exam Tip: For conditional experiments, draw a clear tree diagram to visualize the branching possibilities. Each path from the start to an end node represents a unique outcome in the sample space. Make sure to list the final outcomes correctly based on the paths.

 

Question 12. Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5. B is the event that at least one of the dice shows up a 3. Are the two events A and B (i) mutually exclusive, (ii) exhaustive ? Give arguments in support of your answer.
Answer:
When two dice are rolled, the total sample space S consists of 36 possible outcomes, as listed in Question 3(ii).
S = {(1, 1), (1, 2), ..., (6, 6)}
Let's define the events A and B:
Event A: The sum of the numbers shown on the two dice is 5.
A = {(1, 4), (2, 3), (3, 2), (4, 1)}
Event B: At least one of the dice shows up a 3.
B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3), (6, 3)}
Now we determine if A and B are mutually exclusive and exhaustive:
(i) Mutually Exclusive Events:
Two events are mutually exclusive if they cannot happen at the same time, meaning their intersection is empty (i.e., \( A \cap B = \emptyset \)).
Let's find the intersection of A and B:
\( A \cap B \) = {(2, 3), (3, 2)}
Since \( A \cap B \) = {(2, 3), (3, 2)}, which is not an empty set, events A and B are **not mutually exclusive**. They share common outcomes where the sum is 5 and at least one die shows a 3.
(ii) Exhaustive Events:
Two events are exhaustive if their union covers the entire sample space (i.e., \( A \cup B = S \)).
Let's find the union of A and B:
\( A \cup B \) = {(1, 4), (2, 3), (3, 2), (4, 1), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (4, 3), (5, 3), (6, 3)}
Counting the unique outcomes in \( A \cup B \): There are 13 outcomes.
The total number of outcomes in the sample space S is 36.
Since \( A \cup B \) has 13 outcomes and S has 36 outcomes, \( A \cup B \neq S \).
Therefore, events A and B are **not exhaustive**. There are many outcomes in S that are not included in \( A \cup B \) (e.g., (1,1), (2,2), etc.).
In simple words: When two dice are rolled, event A is getting a sum of 5, and event B is getting at least one 3. They are not mutually exclusive because some outcomes (like (2,3) and (3,2)) are in both events. They are not exhaustive because when you combine all outcomes from A and B, you still don't cover all 36 possible results of rolling two dice.

🎯 Exam Tip: To check for mutually exclusive events, always find their intersection. If the intersection is not empty, they are not mutually exclusive. To check for exhaustive events, find their union. If the union does not equal the entire sample space, they are not exhaustive.