Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 25 Probability

Exercise 25.1

Answer 1.
Answer: As the coin is tossed 800 times, the total number of trials is 800. Let E₁ and E₂ be the events of the coin coming up with a head and a tail respectively.

The number of times E₁ occurs = 415 and the number of times E₂ occurs = 385

\( P(E_1) = \frac{415}{800} = \frac{83}{160} \), and

\( P(E_2) = \frac{385}{800} = \frac{77}{160} \)
In simple words: We count how many times heads came and divide by total tosses. Same for tails. This gives us the probability of each event.

📝 Teacher's Note: Show students that probability is just counting favorable outcomes divided by total outcomes. Use simple numbers first like 5 heads out of 10 tosses.

🎯 Exam Tip: Always write the formula first: P(event) = Number of favorable outcomes ÷ Total number of outcomes. Then substitute values and simplify the fraction.

Answer 2.
Answer: Total number of families = 333 + 392 + 275 = 1000

(i) Number of families having 1 girl = 392
Required probability = \( \frac{\text{Number of families having 1 girl}}{\text{Total number of families}} = \frac{392}{1000} = \frac{49}{125} \)

(ii) Number of families having 2 girls = 275
Required probability = \( \frac{\text{Number of families having 2 girls}}{\text{Total number of families}} = \frac{275}{1000} = \frac{11}{40} \)

(iii) Number of families having no girl = 333
Required probability = \( \frac{\text{Number of families having no girl}}{\text{Total number of families}} = \frac{333}{1000} \)
In simple words: We count families with 1 girl, 2 girls, or no girls. Then divide each count by the total 1000 families. This gives us each probability.

📝 Teacher's Note: Explain that "no girl" means all children are boys. Make students understand that all three probabilities should add up to 1.

🎯 Exam Tip: Always check that your three probabilities add to 1. Write fractions in simplest form. Show the division step clearly.

Answer 3.
Answer: Number of times 2 tails come up = 83
Total number of times the coins were tossed = 300

\( P(\text{2 tails will come up}) = \frac{\text{Number of times 2 tails come up}}{\text{Total number of times the coins were tossed}} = \frac{83}{300} \)

Number of times 1 tail come up = 140
Total number of times the coins were tossed = 300

\( P(\text{1 tail will come up}) = \frac{\text{Number of times 1 tail come up}}{\text{Total number of times the coins were tossed}} = \frac{140}{300} = \frac{7}{15} \)

Number of times no tail come up = 77
Total number of times the coins were tossed = 300

\( P(\text{no tails will come up}) = \frac{\text{Number of times no tail come up}}{\text{Total number of times the coins were tossed}} = \frac{77}{300} \)
In simple words: We tossed two coins together 300 times. We count how many times we got 0 tails, 1 tail, or 2 tails. Then divide each by 300.

📝 Teacher's Note: Use two actual coins and toss them a few times. Show that "no tails" means both coins show heads. "1 tail" means one head and one tail.

🎯 Exam Tip: Read carefully whether it says "1 tail" or "at least 1 tail". These are different. Write all fractions in simplest form.

Answer 4.
Answer: Total number of times die is thrown = 450

(i) Number of times 4 come up on the die = 75

\( P(\text{4 will come up on die}) = \frac{\text{Number of times 4 come up}}{\text{Total number of times the die is thrown}} = \frac{75}{450} = \frac{1}{6} \)

(ii) Number of times less than 4 come up = 73 + 70 + 74 = 217

\( P(<4 \text{ will come up on die}) = \frac{\text{Number of times}(<4) \text{ come up}}{\text{Total number of times the die is thrown}} = \frac{217}{450} \)

(iii) Number of times greater than 4 come up = 80 + 78 = 158

\( P(>4 \text{ will come up on die}) = \frac{\text{Number of times}(>4) \text{ come up}}{\text{Total number of times the die is thrown}} = \frac{158}{450} = \frac{79}{225} \)
In simple words: A die has 6 faces. We count how many times each outcome happened. "Less than 4" means 1, 2, or 3. "Greater than 4" means 5 or 6.

📝 Teacher's Note: Show a real die. Point out that "less than 4" includes 1, 2, and 3. "Greater than 4" includes 5 and 6. Students often forget this.

🎯 Exam Tip: When it says "less than" or "greater than", list all the numbers that satisfy the condition. Then add their frequencies. Don't forget to simplify fractions.

Answer 5.
Answer: The total number of tests taken by a student = 6

(i) The number of times a student gets more than 70% marks in a unit test = 4

\( P(\text{More than 70% marks in a unit test}) = \frac{\text{Number of times student gets more than 70% marks in a unit test}}{\text{Total number of tests taken by a student}} = \frac{4}{6} = \frac{2}{3} \)

(ii) The number of times a student gets less than 72% marks in a unit test = 3

\( P(\text{less than 72% marks in a unit test}) = \frac{\text{Number of times student gets less than 72% marks in a unit test}}{\text{Total number of tests taken by a student}} = \frac{3}{6} = \frac{1}{2} \)

(iii) The number of times a student gets less than 65% marks in a unit test = 0

\( P(\text{less than 65% marks in a unit test}) = \frac{\text{Number of times student gets less than 65% marks in a unit test}}{\text{Total number of tests taken by a student}} = \frac{0}{6} = 0 \)
In simple words: Out of 6 tests, we count how many times the student scored in different ranges. If something never happened, its probability is 0.

📝 Teacher's Note: Explain that probability can be 0 (never happens) or 1 (always happens). Use student marks as it relates to their daily life.

🎯 Exam Tip: When counting marks ranges, be careful about "more than" vs "at least". Zero favorable outcomes gives probability 0. Always simplify fractions.

Answer 7.
Answer: When a die is tossed once, the possible outcomes are the numbers 1, 2, 3, 4, 5, 6

So, total number of possible outcomes = 6

(i) The event is getting an even number and the even numbers are 2, 4, 6.

So, the number of favourable outcomes to the event getting an even number = 3

Therefore, \( P(\text{getting an even number}) = \frac{\text{Favourable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2} \)

(ii) The event is getting a perfect square and the perfect squares are 1 and 4.

So, the number of favourable outcomes to the event getting a perfect square = 2

Therefore, \( P(\text{getting a perfect square}) = \frac{\text{Favourable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3} \)
In simple words: A die has 6 numbers. Even numbers are 2, 4, 6. Perfect squares are numbers like 1×1=1 and 2×2=4. We count these and divide by 6.

📝 Teacher's Note: Show students what perfect squares are: 1=1×1, 4=2×2, 9=3×3. On a die, only 1 and 4 are perfect squares. 9 is too big for a die.

🎯 Exam Tip: List all favorable outcomes clearly before counting them. For perfect squares on a die, only 1 and 4 qualify. Don't include 9 as it's not on a standard die.

Answer 10.
Answer: The sample space for the experiment of tossing two coins is {HH, HT, TH, TT}.

Number of outcomes = 4

Number of favourable outcomes = 2

\( P(\text{getting different faces on the coins}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = \frac{1}{2} \)
In simple words: Two coins can give 4 results: both heads, both tails, or one of each. Different faces means one head and one tail, which happens in 2 ways.

📝 Teacher's Note: Use two different colored coins to show HT and TH are different outcomes. Students often think they are the same.

🎯 Exam Tip: Write all possible outcomes first: HH, HT, TH, TT. Count carefully which ones satisfy the condition. HT and TH both give different faces.

Answer 11.
Answer: When a coin is tossed twice, the outcomes are {HH, HT, TH, TT}.

Number of outcomes = 4

(i) Number of favourable outcomes = 1

\( P(\text{getting no head}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{1}{4} \)

(ii) Number of favourable outcomes = 3

\( P(\text{getting almost one tail}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4} \)
In simple words: "No head" means both coins show tails (TT). "At most one tail" means 0 tails (HH) or 1 tail (HT, TH). That's 3 outcomes out of 4.

📝 Teacher's Note: Explain "at most one" means "one or less". So at most one tail includes zero tails and exactly one tail. Students often miss the zero case.

🎯 Exam Tip: For "at most one tail", include outcomes with 0 tails and 1 tail. Don't forget the case with no tails at all.

Answer 12.
Answer: If three coins are tossed then possible outcomes are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

So total number of outcomes = 8

Favourable outcomes for getting at least 2 heads are:

HHH, HHT, HTH, THH

So number of favourable outcomes = 4

Thus, \( P(\text{Getting at least 2 heads}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2} \)
In simple words: Three coins give 8 possible results. "At least 2 heads" means 2 heads or 3 heads. We count outcomes with 2 or more heads.

📝 Teacher's Note: Use three actual coins. Show that "at least 2 heads" includes exactly 2 heads and exactly 3 heads. Don't forget the case with all 3 heads.

🎯 Exam Tip: Write all 8 outcomes systematically. "At least 2 heads" means 2 or more heads. Count outcomes with exactly 2 heads and exactly 3 heads.

Answer 13.
Answer: If two dice are rolled then the possible outcomes are:

(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)

So the total number of outcomes = 36

(i) a sum of 6

Favourable outcomes for getting a sum 6 are:
(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)

So number of favourable outcomes = 5

Thus, \( P(\text{a sum of 6}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{36} \)

(ii) two different digits

Here we use the following formula:
P(Two different digits) = 1 - P(both digits are same)

Now favourable outcomes for both digits same are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

So the number of possible outcomes for both digits same = 6

Thus, \( P(\text{Two different digits}) = 1 - \frac{6}{36} = \frac{5}{6} \)

(iii) a difference of 1

Favourable outcomes for getting a difference 1 are:
(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6), (6, 5)

So the number of possible outcomes = 10
In simple words: Two dice give 36 outcomes (6×6). For sum of 6, we find pairs that add to 6. For different digits, we subtract same digits from total. For difference of 1, we find pairs where bigger minus smaller equals 1.

📝 Teacher's Note: Draw a 6×6 grid to show all 36 outcomes. Students can see patterns better this way. For difference, explain it's always |first - second| = 1.

🎯 Exam Tip: List outcomes systematically. For difference of 1, check both (a,b) where a-b=1 and b-a=1. Don't miss any pairs.

Answer 14.
Answer: Total number of outcomes = 3

(i) P(Yellow ball) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{1}{3} \)

(ii) P(Red ball) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{1}{3} \)

(iii) P(Blue ball) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{1}{3} \)
In simple words: There are 3 balls total. Each ball has an equal chance of being picked. So each ball has probability \( \frac{1}{3} \).

📝 Teacher's Note: Use 3 different colored balls in class. Let students pick one ball without looking. Each color has the same chance of being picked.

🎯 Exam Tip: Always write the formula first. Then substitute the numbers. Write your final answer as a fraction in simplest form.

Answer 15.
Answer: Since P(Winning) + P(Losing) = 1

Therefore, P(Losing) = 1 - P(winning) = \( 1 - \frac{5}{11} = \frac{6}{11} \)
In simple words: The chance of winning plus the chance of losing must add up to 1 (which means 100%). So we subtract the winning chance from 1 to get the losing chance.

📝 Teacher's Note: Explain that probabilities always add up to 1. Use examples like heads or tails on a coin. If heads is \( \frac{1}{2} \), then tails must also be \( \frac{1}{2} \).

🎯 Exam Tip: Write the formula P(Event) + P(Not Event) = 1 first. Then rearrange to find the unknown probability. Show all steps clearly.

Answer 16.
Answer: Total number of marbles = 5 + 8 + 4 = 17

(i) Number of red marbles = 5

Probability of getting a red marble = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{17} \)

(ii) Number of white marbles = 8

Probability of getting a white marble = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{8}{17} \)

(iii) Number of green marbles = 4

Probability of getting a green marble = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{17} \)

Probability of not getting a green marble = \( 1 - \frac{4}{17} = \frac{13}{17} \)
In simple words: Count all marbles first. Then count how many of each color you want. The probability is the number you want divided by the total number.

📝 Teacher's Note: Use real marbles or colored balls. Let students count each color and the total. This makes the concept very clear and easy to understand.

🎯 Exam Tip: Always find the total first. Write it clearly. For "not getting" questions, use 1 minus the probability of getting that item.

Answer 17.
Answer: Possible outcomes for tossing two coins are: HH, HT, TH, TT

So the number of outcomes = 4

(i) Favourable outcomes are: TT

So, P(Two tails) = \( \frac{1}{4} \)

(ii) Favourable outcomes are: HH

So, P(No tail) = \( \frac{1}{4} \)

(iii) Favourable outcomes are: TT, TH, HT

P(At least one tail) = \( \frac{3}{4} \)
In simple words: When you toss two coins, there are 4 ways they can land. List all possibilities first. Then count how many match what you want.

📝 Teacher's Note: Toss two coins in class many times. Write down all results on the board. Students will see the pattern of 4 possible outcomes clearly.

🎯 Exam Tip: Always list all possible outcomes first. Then circle the favourable ones. This prevents counting mistakes. For "at least one" questions, count all except "none".

Answer 18.
Answer: Total number of outcomes when two dice are rolled = 36

(i) A doublet

Favourable outcomes are: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

Number of favourable outcomes = 6

P(A doublet) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6} \)

(ii) Sum divisible by 5

Favourable outcomes are: (1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4)

Number of favourable outcomes = 7

P(Sum divisible by 5) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{7}{36} \)

(iii) A sum of at least 11

Favourable outcomes are: (5,6), (6,5), (6,6)

Number of favourable outcomes = 3

P(sum is at least 11) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12} \)
In simple words: Two dice give 36 total outcomes (6 × 6). A doublet means both dice show the same number. For sums, add the numbers on both dice.

📝 Teacher's Note: Draw a 6×6 grid on the board. Fill in all 36 outcomes. Students can easily see and count the favourable outcomes for any question.

🎯 Exam Tip: Remember that two dice give 36 total outcomes. List all favourable outcomes carefully. For "at least 11", the sums are 11 or 12 only.

Answer 19.
Answer: When two dice are rolled the outcomes are:

(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)

So, the total number of outcomes = 36
In simple words: This shows all 36 ways two dice can land. The first number is from the first die, the second number is from the second die.

📝 Teacher's Note: This is the complete sample space for two dice. Students should memorize that two dice always give 36 total outcomes. Use this list for all dice probability problems.

🎯 Exam Tip: You don't need to write all 36 outcomes in exams. Just write "Total outcomes = 36" and list only the favourable ones for your question.