CBSE Class 10 Mathematics Statistics VBQs Set 05

Question. Two unbiased coins are tossed simultaneously then the probability of getting no head is \( \frac{A}{B} \), then \( (A+B)^2 \) is? 
Answer: If two unbiased coins are tossed simultaneously we obtained possible outcomes:
HH, HT, TH, TT
Hence, total number of outcomes = 4
No head is obtained if the event TT occurs.
Hence, number of favourable outcomes = 1
Hence, required probability = \( \frac{1}{4} \)
But, given probability = \( \frac{A}{B} \)
So, A = 1 and B = 4
therefore, \( (A+B)^2 = (1+4)^2 = (5)^2 = 25 \)
So, \( (A+B)^2 = 25 \).

Question. If odds in against of an event be 3 : 8, then the probability of occurrence of this event is?
Answer: Let the event be E.
Number of favourable outcomes to event E are \( m \) and total outcomes be \( n \).
According to the question,
\( \Rightarrow \frac{m}{n-m} = \frac{3}{8} \)
\( \Rightarrow 8m = 3n - 3m \)
\( \Rightarrow 11m = 3n \)
\( \frac{m}{n} = \frac{3}{11} \)
Hence, \( P(E) = \frac{\text{number of outcomes favourable to E}}{\text{number of total outcomes}} = \frac{m}{n} = \frac{3}{11} \)
So, probability of occurrence of this event is \( \frac{3}{11} \).

Question. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is \( \frac{1}{4} \). The probability of selecting a blue ball at random from the same jar is \( \frac{1}{3} \). If the jar contains 10 orange balls, find the total number of balls in the jar.
Answer: \( P(\text{Red}) = \frac{1}{4} \), \( P(\text{blue}) = \frac{1}{3} \)
\( P(\text{orange}) = 1 - \frac{1}{4} - \frac{1}{3} = \frac{5}{12} \)
\( \frac{5}{12} \times (\text{Total no. of balls}) = 10 \)
Total no. of balls = \( \frac{10 \times 12}{5} = 24 \) 

Question. The following table gives the number of participants in a yoga camp:

Find the modal age of the participants. 
Answer: In the given frequency distribution the modal class is 50-60 having maximum frequency as 90.
\( \text{Mode, } M_o = l + \left( \frac{f - f_1}{2f - f_1 - f_2} \right) \times h \)
Here, the lower limit of the modal class, \( l = 50 \)
Frequency of the modal class, \( f = 90 \)
Frequency of the class preceding modal class, \( f_1 = 58 \)
Frequency of the class succeeding modal class, \( f_2 = 83 \)
Height of the class mark, \( h = 10 \)
\( M_o = 50 + \left( \frac{90 - 58}{180 - 58 - 83} \right) \times 10 \)
\( = 50 + \frac{32}{180 - 141} \times 10 \)
\( = 50 + \frac{32}{39} \times 10 \)
\( = 50 + 8.21 \)
\( = 58.2 \)
Hence, the modal age of the given distribution is 58.2 years.

Question. Calculate the mean of the following frequency distribution:

Answer:

Now, \( \text{mean } \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3280}{50} = 65.6 \)
Hence, the mean of the given distribution is 65.6.

Question. Two different dice are thrown at the same time. Find the probability that the number appearing on the two dice
(A) Have a sum 8.
(B) Are first even and second odd. 
Answer: When 2 dice are thrown, then 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)
Total number of outcomes = 36
(A) Let E be the event that the numbers appearing on the dice have a sum 8.
Then, favourable cases = (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
\( \therefore \) Total number of favourable cases = 5
Now, \( P(E) = \frac{5}{36} \)
Hence, the required probability is \( \frac{5}{36} \).
(B) Let A be the event that the number appearing on first dice is even and second is odd.
Then, favourable outcomes = (2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)
\( \therefore \) Total number of favourable outcomes = 9
Now, \( P(A) = \frac{9}{36} = \frac{1}{4} \)
Hence, the required probability is \( \frac{1}{4} \).

Question. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(A) a two-digit number, (B) a number divisible by 5, (C) perfect square number. 
Answer: Total number of outcomes = 90
(A) Let E be the event that a disc drawn bear a two-digit number.
Favourable cases are 10, 11, 12, 13..., 90
Total number of favourable outcomes = 81
\( P(E) = \frac{81}{90} = \frac{9}{10} \)
Hence, the required probability is \( \frac{9}{10} \).
(B) Let \( E_1 \) be the event that a disc drawn bears a number divisible by 5.
Then favourable cases are 5, 10, 15, 20, 25 ..., 90.
Total number of favourable outcomes = 18
\( P(E_1) = \frac{18}{90} = \frac{1}{5} \)
Hence, the required probability is \( \frac{1}{5} \).
(C) Let \( E_2 \) be the event that a card drawn bears a perfect square number.
Then, number having perfect square are: 1, 4, 9, 16, 25, 36, 49, 64, 81
Total number of favourable outcomes = 9
\( P(E_2) = \frac{9}{90} = \frac{1}{10} \)
Hence, the probability that the selected card bears a perfect square number is \( \frac{1}{10} \).

Question. Peter throws two different dice together and find the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25. 
Answer: When 2 different dice are thrown by Peter, 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)
Total number of outcomes = 36
The favourable outcomes for getting the product of numbers on the dice equal to 25 is (5, 5)
Favourable number of outcomes = 1
\( \therefore \) (Peter gets the product of number as 25) = \( \frac{1}{36} \) ...(i)
When 1 dice is thrown by Rina, then possible outcomes are: 1, 2, 3, 4, 5, 6.
\( \therefore \) Total number of outcomes = 6
Rina throws a die and squares the number, so to get the number 25, the favourable outcome is 5.
\( \therefore \) P(Rina gets the product of number as 25) = \( \frac{1}{6} \) ...(ii)
From (i) and (ii), we get that: \( P(\text{Rina}) > P(\text{Peter}) \)
Hence, Rina has better chance to get the number 25.

Question. A die is thrown twice. Find the probability that:
(A) 5 will not come either time.
(B) The sum of numbers on the two dice is not more than 5. 
Answer: When a dice is thrown twice. Then the total 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)
Then, the total number of outcomes = 36
(A) Outcomes when 5 comes up = (1, 5), (2, 5), (3, 5), (4, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
Total outcomes when 5 comes up = 11
Total outcomes when 5 will not come up = 36 - 11 = 25
\( \therefore P(\text{that 5 will not come up either times}) = \frac{25}{36} \)
(B) Outcomes when the sum is not more than 5 = (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)
Total outcomes when sum is not more than 5 = 10
\( P(\text{that sum is not more than 5}) = \frac{10}{36} = \frac{5}{18} \)
Hence, the required probability is \( \frac{5}{18} \).

Question. A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing to one of the numbers 1, 2, 3, ...., 8 which are equally likely outcomes. What is the probability that the arrow will point at (A) an odd number (B) a number greater than 3 (C) a number less than 9. 
Answer: Total numbers = 8
(A) Odd numbers that can be outcome = 1, 3, 5, 7
Number of odd number = 4
\( P(\text{getting an odd number}) = \frac{\text{Number of odd numbers}}{\text{Total numbers}} = \frac{4}{8} = \frac{1}{2} \)
Hence, the required probability is \( \frac{1}{2} \).
(B) Numbers greater than 3 are = 4, 5, 6, 7, 8
Number of numbers greater than 3 = 5
\( P(\text{getting a number greater than 3}) = \frac{5}{8} \)
Hence, the required probability is \( \frac{5}{8} \).
(C) Number less than 9 = 1, 2, 3, 4, 5, 6, 7, 8
Number of numbers less than 9 = 8
\( P(\text{getting a number less than 9}) = \frac{8}{8} = 1 \)
Hence, the required probability is 1.

Question. Two different dice are thrown together. Find the probability that the numbers obtained have (i) even sum, and (ii) even product.
Answer: i) \( A = \text{sum of digits is even} \).
\( n(S) = 6^2 = 36 \). \( - \text{total possible outcomes} \).
\( n(A) = \{ (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6) \} \)
\( = 18 \).
\( P(A) = \frac{n(A)}{n(S)} = \frac{18}{36} = \frac{1}{2} \text{ or } 0.5 \).
\( \therefore \text{probability of getting an even sum is } \frac{1}{2} \text{ or } 0.5 \).
ii) \( A = \text{product of digits is even} \).
\( n(S) = 36 \).
\( n(A) = \{ (1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \} \)
\( = 27 \).
\( P(A) = \frac{n(A)}{n(S)} = \frac{27}{36} = \frac{3}{4} = 0.75 \).
\( \therefore \text{probability of getting even product is } \frac{3}{4} \text{ or } 0.75 \). 

Question. A number \( x \) is selected at random from the numbers 1, 2, 3 and 4. Another number \( y \) is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of \( x \) and \( y \) is less than 16.
Answer: Total possible outcome = 1, 2, 3, 4 & 1, 4, 9, 16 = 16
\( \begin{array}{|c|c|c|c|c|} \hline x \backslash y & 1 & 4 & 9 & 16 \\ \hline 1 & 1 & 4 & 9 & 16 \\ \hline 2 & 2 & 8 & 18 & 32 \\ \hline 3 & 3 & 12 & 27 & 48 \\ \hline 4 & 4 & 16 & 36 & 64 \\ \hline \end{array} \)
Total favourable event having product less than 16
\( = \{1, 4, 9, 2, 8, 3, 12, 4\} \text{ Count } = 8 \).
\( \text{Probability} = \frac{\text{Favourable event outcome}}{\text{Total event}} \)
\( P[E] = \frac{8}{16} = \frac{1}{2} \).
\( \text{Ans: } \frac{1}{2} \) [

Question. A number \( x \) is selected from the numbers 1, 2, 3 and then a second number \( y \) is selected randomly from the numbers 1, 4, 9. What is the probability that the product \( xy \) of the two numbers will be less than 9?
Answer: Total number of products is 9, say {1, 4, 9, 2, 8, 18, 3, 12, 27}.
Of these five are less than 9.
So, \( P(xy < 9) = \frac{5}{9} \)

Question. A bag contains 24 balls of which \( x \) are red, \( 2x \) are white and \( 3x \) are blue. A ball is drawn at random. What is the probability that it is: (A) not a red ball? (B) a white ball? (C) either a blue or a white ball?
Answer: Total balls = \( x + 2x + 3x = 6x \)
(A) \( P(\text{not a red ball}) = \frac{2x + 3x}{6x} = \frac{5}{6} \)
(B) \( P(\text{a white ball}) = \frac{2x}{6x} = \frac{1}{3} \)
(C) \( P(\text{a blue or a white}) = P(\text{not a red ball}) = \frac{5}{6} \) [by (a) above]

LONG ANSWER Type Questions

Question. Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately. 
Answer: First dice {1, 2, 3, 4, 5, 6}
Second dice {1, 1, 2, 2, 3, 3}
Total number of outcomes, \( T(E) = 36 \).
(1) Let \( E_1 \) be the event of getting sum 2.
Favourable outcomes are (1, 1), (1, 1)
No. of favourable outcomes, \( F(E_1) = 2 \)
\( \therefore \) Required probability \( P(E_1) = \frac{F(E_1)}{T(E)} = \frac{2}{36} = \frac{1}{18} \)
Hence, \( P(E_1) = \frac{1}{18} \).
(2) Let \( E_2 \) be the event of getting sum 3.
Favourable outcomes are (1, 2), (1, 2), (2, 1), (2, 1)
\( \therefore \) No. of favourable outcomes, \( F(E_2) = 4 \)
\( \therefore \) Required probability \( P(E_2) = \frac{4}{36} = \frac{1}{9} \)
Hence, \( P(E_2) = \frac{1}{9} \).
(3) Let \( E_3 \) be the event of getting sum 4.
Favourable outcomes are (1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)
\( \therefore \) No. of favourable outcomes \( F(E_3) = 6 \)
\( \therefore \) Required probability \( P(E_3) = \frac{6}{36} = \frac{1}{6} \)
Hence, \( P(E_3) = \frac{1}{6} \).
(4) Let \( E_4 \) be the event of getting sum 5.
Favourable outcomes are (2, 3), (2, 3), (4, 1), (4, 1), (3, 2), (3, 2)
No. of favourable outcomes \( F(E_4) = 6 \).
Required probability \( P(E_4) = \frac{6}{36} = \frac{1}{6} \)
Hence, \( P(E_4) = \frac{1}{6} \).
(5) Let \( E_5 \) be the event of getting sum 6.
Favourable outcomes are (3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)
No. of favourable outcomes \( F(E_5) = 6 \)
Required Probability \( P(E_5) = \frac{6}{36} = \frac{1}{6} \)
Hence, \( P(E_5) = \frac{1}{6} \).
(6) Let \( E_6 \) be the event of getting sum 7.
Favourable outcomes are (4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)
No. of favourable outcomes \( F(E_6) = 6 \)
Required probability \( P(E_6) = \frac{6}{36} = \frac{1}{6} \)
Hence, \( P(E_6) = \frac{1}{6} \).
(7) Let \( E_7 \) be the event of getting sum 8.
Favourable outcomes are (5, 3), (5, 3), (5, 3), (6, 2), (6, 2)
No. of favourable outcomes \( F(E_7) = 4 \)
Required probability \( P(E_7) = \frac{4}{36} = \frac{1}{9} \)
Hence, \( P(E_7) = \frac{1}{9} \).
(8) Let \( E_8 \) be the event of getting sum 9.
Favourable outcomes are (6, 3), (6, 3)
No. of favourable outcomes, \( F(E_8) = 2 \)
\( \therefore \) Required Probability \( P(E_8) = \frac{2}{36} = \frac{1}{18} \)
Hence, \( P(E_8) = \frac{1}{18} \).
Hence, \( P(E_1) = \frac{1}{18}, P(E_2) = \frac{1}{9}, P(E_3) = \frac{1}{6}, P(E_4) = \frac{1}{6}, P(E_5) = \frac{1}{6}, P(E_6) = \frac{1}{6}, P(E_7) = \frac{1}{9}, P(E_8) = \frac{1}{18} \).

Question. The average score of boys in the examination of a school is 71 and that of the girls is 73. The average score of the school in the examination is 71.8. Find the ratio of the number of boys to the number of girls who appeared in the examination. 
Answer: Let the number of boys in the school be \( x \) and the number of girls in the school be \( y \).
Average score of boys = 71
Total score of boys in the examination of school = \( 71x \)
Average score of girls = 73
Total score of girls in the examination of school = \( 73y \)
Average score of the school in examination = 71.8
\( \frac{\text{total score of boys + total score of girls}}{\text{total score of boys and girls}} = 71.8 \)
\( \Rightarrow \frac{71x + 73y}{x + y} = 71.8 \)
\( \Rightarrow 71x + 73y = 71.8x + 71.8y \)
\( \Rightarrow 73y - 71.8y = 71.8x - 71x \)
\( \Rightarrow 1.2y = 0.8x \)
\( \Rightarrow \frac{1.2}{0.8} = \frac{x}{y} \)
\( \Rightarrow \frac{12}{8} = \frac{x}{y} \)
\( \Rightarrow \frac{x}{y} = \frac{3}{2} \)
\( x : y = 3 : 2 \)