GSEB Class 10 Maths Solutions Chapter 15 Probability Exercise 15.2

Gujarat Board Textbook Solutions Class 10 Maths Chapter 15 Probability Ex 15.2

 

Question 1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on:
(i) the same day?
(ii) consecutive days?
(iii) different days?
Answer:
There are 5 weekdays from Tuesday to Saturday. Each customer can visit the shop on any of these 5 days. So, the total number of possible ways for both Shyam and Ekta to visit the shop is \( 5 \times 5 = 25 \). This is our total sample space, \( n(S) = 25 \).
(i) For them to visit on the same day, the favorable outcomes are:
(Tuesday, Tuesday), (Wednesday, Wednesday), (Thursday, Thursday), (Friday, Friday), (Saturday, Saturday).
Let A be the event that they visit on the same day. The number of favorable outcomes for A is \( n(A) = 5 \).
The probability of event A is \( P(A) = \frac {n(A)}{n(S)} = \frac {5}{25} = \frac {1}{5} \).
(ii) For them to visit on consecutive days, the favorable outcomes are:
(Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday)
(Wednesday, Tuesday), (Thursday, Wednesday), (Friday, Thursday), (Saturday, Friday)
Let B be the event that they visit on consecutive days. The number of favorable outcomes for B is \( n(B) = 8 \).
The probability of event B is \( P(B) = \frac {n(B)}{n(S)} = \frac {8}{25} \).
(iii) For them to visit on different days, we can subtract the outcomes where they visit on the same day from the total number of outcomes.
Total outcomes = 25
Outcomes for visiting on the same day = 5
So, the number of favorable outcomes for visiting on different days = \( 25 - 5 = 20 \).
Let C be the event that they visit on different days. The number of favorable outcomes for C is \( n(C) = 20 \).
The probability of event C is \( P(C) = \frac {n(C)}{n(S)} = \frac {20}{25} = \frac {4}{5} \).
In simple words: First, count all the different ways Shyam and Ekta can visit the shop (5x5=25). For part (i), find how many times they visit on the identical day (5 ways) and divide by 25. For part (ii), count how many ways they visit one day after the other (8 ways) and divide by 25. For part (iii), subtract the "same day" visits from the total visits (25-5=20 ways) and divide by 25.

Exam Tip: When calculating probabilities for multiple conditions (same day, consecutive days, different days), always ensure that your total sample space \( n(S) \) remains consistent for all calculations within the same question.

 

Question 2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

What is the probability that the total score is (i) even? (ii) 6? (iii) at least 6?
Answer:
First, we need to complete the table of all possible sums when two dice are thrown. The numbers on the faces are 1, 2, 2, 3, 3, 6.
The number of possible outcomes, \( n(S) \), is 36.
(i) Let A be the event that the total score is an even number. Counting the even numbers from the table, we get 18 even scores.
So, \( n(A) = 18 \).
The probability \( P(A) = \frac {n(A)}{n(S)} = \frac {18}{36} = \frac {1}{2} \).
(ii) Let B be the event that the total score is 6. Counting the occurrences of 6 in the table, we find 4 scores of 6.
So, \( n(B) = 4 \).
The probability \( P(B) = \frac {n(B)}{n(S)} = \frac {4}{36} = \frac {1}{9} \).
(iii) Let C be the event that the total score is at least 6 (meaning 6 or greater). Counting all scores \( \ge 6 \) from the table, we find 15 such scores.
So, \( n(C) = 15 \).
The probability \( P(C) = \frac {n(C)}{n(S)} = \frac {15}{36} = \frac {5}{12} \).
In simple words: First, fill out the entire table to see all possible sums, which total 36. For part (i), count how many of these sums are even and divide by 36. For part (ii), count how many sums are exactly 6 and divide by 36. For part (iii), count how many sums are 6 or higher and divide by 36.

Exam Tip: Always construct the complete sample space (like the sum table) when dealing with two dice or multiple events to accurately count favorable outcomes for each part of the question.

 

Question 3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red ball, find the number of balls in the bag. (National Olympiad)
Answer:
Number of red balls in the bag = 5
Let the number of blue balls in the bag = \( x \)
The total number of balls in the bag, \( n(S) = x + 5 \).
Let A be the event of drawing red balls. Then, the number of favorable outcomes for A is \( n(A) = 5 \).
The probability of getting a red ball, \( P(A) = \frac {n(A)}{n(S)} = \frac {5}{x + 5} \).
Let B be the event of drawing blue balls. Then, the number of favorable outcomes for B is \( n(B) = x \).
The probability of getting a blue ball, \( P(B) = \frac {n(B)}{n(S)} = \frac {x}{x + 5} \).
According to the question, the probability of drawing a blue ball is double that of a red ball:
\( P(B) = 2 \times P(A) \)

\( \implies \frac {x}{x + 5} = 2 \times \frac {5}{x + 5} \)

\( \implies \frac {x}{x + 5} = \frac {10}{x + 5} \)
Since the denominators are equal and non-zero (as \( x+5 \) cannot be zero), the numerators must also be equal:
\( \implies x = 10 \)
So, the number of blue balls in the bag is 10.
The total number of balls in the bag = Number of red balls + Number of blue balls = \( 5 + 10 = 15 \).
In simple words: We have 5 red balls and 'x' blue balls, making a total of \( x+5 \) balls. The chance of picking a red ball is \( \frac{5}{x+5} \), and for a blue ball is \( \frac{x}{x+5} \). The problem says the blue ball chance is twice the red ball chance. Set up the equation \( \frac{x}{x+5} = 2 \times \frac{5}{x+5} \) and solve for 'x'. You will find that 'x' equals 10.

Exam Tip: Define variables clearly for unknown quantities. Formulate the probabilities based on these variables and the total outcomes. Then, use the given relationship between the probabilities to create an equation and solve for the unknown.

 

Question 4. A box contains 12 balls out of which x are black.
1. If one ball is drawn at random, what is the probability that it will be a black ball?
2. If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before, find x.
Answer:
Total number of balls in the box = 12
Number of black balls = \( x \)
1. The probability of drawing a black ball initially, \( P_1 \), is:
\( P_1 = \frac {\text{Number of black balls}}{\text{Total number of balls}} = \frac {x}{12} \)
2. If 6 more black balls are added to the box:
New number of black balls = \( x + 6 \)
New total number of balls = \( 12 + 6 = 18 \)
The new probability of drawing a black ball, \( P_2 \), is:
\( P_2 = \frac {x + 6}{18} \)
According to the question, the new probability \( P_2 \) is double the initial probability \( P_1 \):
\( P_2 = 2 \times P_1 \)

\( \implies \frac {x + 6}{18} = 2 \times \frac {x}{12} \)

\( \implies \frac {x + 6}{18} = \frac {2x}{12} \)

\( \implies \frac {x + 6}{18} = \frac {x}{6} \)
To solve this equation, cross-multiply:
\( \implies 6(x + 6) = 18x \)
\( \implies 6x + 36 = 18x \)
Subtract \( 6x \) from both sides:
\( \implies 36 = 18x - 6x \)
\( \implies 36 = 12x \)
Divide by 12 to find \( x \):
\( \implies x = \frac {36}{12} \)
\( \implies x = 3 \)
So, initially there were 3 black balls.
In simple words: First, calculate the chance of drawing a black ball from 12 balls when 'x' are black. Next, add 6 more black balls, which changes the number of black balls to \( x+6 \) and the total balls to 18. Then, calculate the new chance of drawing a black ball. The problem states this new chance is twice the original chance. Form an equation with these probabilities and solve for 'x'.

Exam Tip: Clearly distinguish between the initial and changed conditions (number of balls, total balls, probabilities) to avoid confusion in setting up the equations.

 

Question 5. A jar contains 24 marbles, some are green, and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \( \frac {2}{3} \) Find the number of blue marbles in the jar.
Answer:
Total number of marbles in the jar = 24
Let the number of green marbles = \( x \)
Then, the number of blue marbles = \( 24 - x \)
The probability of drawing a green marble is given as \( \frac {2}{3} \).
We also know that probability is calculated as the number of favorable outcomes divided by the total number of outcomes.
So, \( P(\text{green marble}) = \frac {\text{Number of green marbles}}{\text{Total number of marbles}} = \frac {x}{24} \)
According to the problem, this probability equals \( \frac {2}{3} \):
\( \frac {x}{24} = \frac {2}{3} \)
To find \( x \), we can multiply both sides of the equation by 24:
\( x = \frac {2}{3} \times 24 \)
\( x = 2 \times 8 \)
\( x = 16 \)
So, the number of green marbles is 16.
Now, we need to find the number of blue marbles:
Number of blue marbles = Total marbles - Number of green marbles
Number of blue marbles = \( 24 - 16 = 8 \).
Therefore, there are 8 blue marbles in the jar.
In simple words: There are 24 marbles in total. Let 'x' be the green marbles, so \( 24-x \) are blue. The chance of picking a green marble is shown as \( \frac{x}{24} \), and it is also given as \( \frac{2}{3} \). Set \( \frac{x}{24} = \frac{2}{3} \) and solve for 'x'. Once you find 'x' (the number of green marbles), subtract it from the total (24) to get the number of blue marbles.

Exam Tip: Clearly define the variables and use the given probability to form a simple equation. Remember to solve for the final quantity asked in the question, which might not be the variable you initially solved for.