RS Aggarwal Class 7 Mathematics Solutions Chapter 23 Probability

 

Question 1. A coins is tossed 300 times and we get head = 136 times and tail = 164 times. When a coin is tossed at random, what is the probability of getting (i) a head (ii) a tail ?
Answer: (i) The total number of coin tosses performed was 300. Heads appeared 136 times. Using the probability formula, we get \( P(E) = \frac{136}{300} = \frac{34}{75} \).

(ii) The total number of coin tosses performed was 300. Tails appeared 164 times. Using the probability formula, we get \( P(E) = \frac{164}{300} = \frac{41}{75} \).
In simple words: Probability means how often something happens. Divide the number of times an event occurred by the total number of tries. For heads, it's 136 out of 300; for tails, it's 164 out of 300.

Exam Tip: Always simplify fractions to their lowest form by finding the greatest common divisor. The sum of all probabilities in a trial must equal 1 - verify this as a check.

 

Question 2. Two coins are tossed on simultaneously 200 times and we get two heads = 58 times ; one head = 83 times, 0 head = 59 times. When two coins are tossed at random, what is the probability of getting (i) 2 heads (ii) 1 head (iii) 0 head ?
Answer: The two coins were tossed together a total of 200 times. The outcomes were recorded as: two heads occurred 58 times, one head occurred 83 times, and no heads occurred 59 times.

(i) Probability of getting 2 heads: \( P(E) = \frac{58}{200} = \frac{29}{100} \)

(ii) Probability of getting one head: \( P(E) = \frac{83}{200} \)

(iii) Probability of getting no head: \( P(E) = \frac{59}{200} \)
In simple words: When you toss two coins, you can get two heads, one head, or no heads. Count how many times each happened out of 200 tosses, then divide to find the probability for each outcome.

Exam Tip: Check that all three probabilities add up to 1 (or 200/200 before simplifying). If they don't sum to 1, you have made an error.

 

Question 3. A dice is thrown 100 times and the out comes are noted as given below:

When a dice is thrown at random, what is the probability of getting a (i) 3 (ii) 6 (iii) 4 (iv) 1 ?
Answer: The dice was rolled 100 times in total.

(i) The number 3 appeared 18 times. Therefore, \( P(E) = \frac{18}{100} = \frac{9}{50} \)

(ii) The number 6 appeared 9 times. Therefore, \( P(E) = \frac{9}{100} \)

(iii) The number 4 appeared 15 times. Therefore, \( P(E) = \frac{15}{100} = \frac{3}{20} \)

(iv) The number 1 appeared 21 times. Therefore, \( P(E) = \frac{21}{100} \)
In simple words: For each outcome on the dice, count how many times it showed up. Divide that count by 100 (the total rolls) to find the probability of rolling that number.

Exam Tip: Always read the frequency table carefully. The sum of all frequencies should match the total number of trials (here, 21 + 14 + 18 + 15 + 23 + 9 = 100).

 

Question 4. In a survey of 100 ladies, it was found that 36 like coffee while 64 dislike it. Out of these ladies, one is chosen at random. What is the probability that the chosen lady (i) likes coffee (ii) dislikes coffee ?
Answer: A survey included 100 ladies in total. Among them, 36 enjoyed coffee and 64 did not enjoy coffee.

(i) Probability that the chosen lady likes coffee: \( P(E) = \frac{36}{100} = \frac{9}{25} \)

(ii) Probability that the chosen lady dislikes coffee: \( P(E) = \frac{64}{100} = \frac{16}{25} \)
In simple words: Out of 100 ladies, 36 like coffee. So the chance of picking a lady who likes coffee is 36 out of 100. The remaining 64 dislike it, so the chance of picking a lady who dislikes it is 64 out of 100.

Exam Tip: The two probabilities must add up to 1 since every lady either likes or dislikes coffee - there is no middle ground. Verify: 9/25 + 16/25 = 25/25 = 1.