Chapter 13 · Statistics and Probability

NCERT Exemplar Class 10 Maths Exercise 13.2

All 14 questions of NCERT Exemplar Class 10 Maths Exercise 13.2 (Statistics and Probability), in the book's order, each with its answer and a step-by-step solution.

1. Q1
   The median of an ​ungrouped data and the ​median calculated when the same data is grouped are always the same. ​Do you ​think that ​this is ​a correct ​statement? ​Give reason.
   Show answer

   No, the statement is not correct. The grouped median uses an interpolation that assumes uniform spread within the median class, so it is only an estimate and generally differs from the exact ungrouped median.

   Step-by-step solution →
2. Q2
   In ​calculating ​the mean of grouped data, grouped in classes of equal ​width, we ​may use the ​formula $\bar{x}=a+\frac{\sum f_i{d}_i}{\sum f_i}$ where $a$ is the assumed mean. $a$ must be one of ​the mid-points of the classes. Is the ​last statement correct? ​Justify your answer.
   Show answer

   No, the statement is not correct. The assumed mean $a$ can be any convenient number; it need not be one of the class mid-points, because $a$ cancels out and the final mean is unchanged.

   Step-by-step solution →
3. Q3
   Is it ​true ​to say that the mean, ​mode and ​median ​of grouped data ​will ​always be ​different? Justify your answer.
   Show answer

   No, it is not true. For a symmetric grouped distribution the mean, median and mode are all equal, so they need not always be different.

   Step-by-step solution →
4. Q4
   Will the ​median class and modal ​class of ​grouped ​data ​always be ​different? Justify your answer.
   Show answer

   No, they need not always be different. For the distribution with frequencies 2, 8, 2 over classes 0-10, 10-20, 20-30, the class 10-20 is both the modal class and the median class.

   Step-by-step solution →
5. Q5
   In a family having three children, ​there may ​be no girl, one girl, two ​girls or ​three girls. So, the ​probability of each ​is $\frac{1}{4}$. ​Is this correct? Justify your answer.
   Show answer

   No, it is not correct. The four events are not equally likely; $P(0)=\frac{1}{8}$, $P(1)=\frac{3}{8}$, $P(2)=\frac{3}{8}$, $P(3)=\frac{1}{8}$.

   Step-by-step solution →
6. Q6
   A game consists of spinning an ​arrow which ​comes to rest ​pointing at ​one of the regions (1, ​2 or 3) (Fig. 13.1). Are the ​outcomes 1, 2 and 3 equally ​likely to occur? ​Give reasons.

   Show answer

   No. Regions 1 and 2 are quarter-circles ($P=\frac{1}{4}$ each) while region 3 is a semicircle ($P=\frac{1}{2}$), so outcome 3 is twice as likely; the outcomes are not equally likely.

   Step-by-step solution →
7. Q7
   Apoorv throws two ​dice once and computes the product of the numbers ​appearing on the ​dice. Peehu throws ​one die and squares the number that ​appears on ​it. Who ​has the better chance of getting the number ​36? Why?
   Show answer

   Peehu has the better chance. $P_{Peehu}=\frac{1}{6}$ while $P_{Apoorv}=\frac{1}{36}$.

   Step-by-step solution →
8. Q8
   When we toss a ​coin, ​there are two ​possible ​outcomes - Head or Tail. Therefore, ​the probability of ​each outcome is $\frac{1}{2}$. Justify ​your answer.
   Show answer

   Yes, the statement is correct provided the coin is fair: Head and Tail are then equally likely, so each has probability $\frac{1}{2}$.

   Step-by-step solution →
9. Q9
   A student says that if you throw a die, it will show up 1 or not ​1. ​Therefore, the probability of ​getting 1 and the probability of getting 'not ​1' each is equal to $\frac{1}{2}$. Is this correct? ​Give reasons.
   Show answer

   No, the student is incorrect. $P(1)=\frac{1}{6}$ and $P(\text{not }1)=\frac{5}{6}$; the two events are not equally likely.

   Step-by-step solution →
10. Q10
    I toss ​three ​coins together. ​The possible ​outcomes are no heads, 1 head, 2 heads and 3 heads. So, I say ​that probability of no ​heads is $\frac{1}{4}$. What ​is wrong with this conclusion?
    Show answer

    The four listed events are not equally likely, so dividing by 4 is wrong; the correct probability of no heads is $\frac{1}{8}$.

    Step-by-step solution →
11. Q11
    If you toss a coin ​6 times and it comes ​down heads on each occasion, ​can ​you say that the ​probability ​of getting ​a head is 1? Give reasons.
    Show answer

    No. Getting 6 heads is just one chance outcome (probability $1/64$, not impossible). The probability of a head for a fair coin stays $1/2$; only a very large number of trials would make the experimental ratio approach $1/2$.

    Step-by-step solution →
12. Q12
    Sushma ​tosses a ​coin 3 times and gets tail each time. Do you think that the ​outcome ​of next ​toss will ​be ​a tail? Give reasons.
    Show answer

    No. Each toss is independent, so the next toss is still equally likely to be a head or a tail, each with probability $\frac{1}{2}$.

    Step-by-step solution →
13. Q13
    If I toss a ​coin 3 times ​and get head each ​time, should I expect a tail to have a ​higher chance in the ​4th toss? ​Give reason ​in ​support of your answer.
    Show answer

    No - the 4th toss is independent of the first three, so a tail still has probability $\frac{1}{2}$.

    Step-by-step solution →
14. Q14
    A bag contains slips numbered from 1 to 100. ​If Fatima chooses a slip at ​random ​from the ​bag, it will either be an ​odd number or an even number. Since ​this situation has only two possible outcomes, ​so, the probability of each is $\frac{1}{2}$. Justify.
    Show answer

    Yes, each probability is $\frac{1}{2}$, but the justification 'two outcomes so 1/2 each' is wrong. It is 1/2 because there are exactly 50 odd and 50 even numbers among the 100 slips.

    Step-by-step solution →