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SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 10 Mathematics Statistics and Probability

96 questions · 96 still being checked

EXERCISE 13.2 11–14 (part 5 of 11)

  1. Exercise 11

    If you toss a coin 6\displaystyle 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1\displaystyle 1? Give reasons.

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    NCERT’s answer
    No, the outcomes 'head' and 'tail' are equally likely every time regardless of what you get in a few tosses.
    On every toss, Head and Tail are equally likely. \[P(\text{Head}) = \frac{1}{2} \quad \text{on each toss} \] Six heads in a row do not change this. Answer: No. \(\displaystyle P(\text{Head})=\tfrac12\), not \(\displaystyle 1\).

  2. Exercise 12

    Sushma tosses a coin 3\displaystyle 3 times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons.

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    NCERT’s answer
    It could be a tail or head as both the outcomes are equally likely, in each toss.
    Each toss of a fair coin is an independent trial; the coin has no memory of the outcomes \(\displaystyle T,T,T\) already seen. \[P(\text{Tail on toss 4}) = \frac{1}{2} \] This probability is unchanged by the preceding three tails. Answer: No — the 4th toss remains equally likely to be Head or Tail, \(\displaystyle P(\text{Tail})=\tfrac12\).

  3. Exercise 13

    If I toss a coin 3\displaystyle 3 times and get head each time, should I expect a tail to have a higher chance in the 4th \displaystyle 4^{\text {th }} toss? Give reason in support of your answer.

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    NCERT’s answer
    No, head and tail are equally likely. So, no question of expecting a tail to have a higher chance in the 4th toss.
    Successive tosses of a fair coin are independent events, so the three earlier heads exert no influence on the next toss. \[P(\text{Head on toss 4}) = P(\text{Tail on toss 4}) = \frac{1}{2} \] Expecting a tail "to balance the run" is the gambler's fallacy — the coin does not compensate for past outcomes. Answer: No — both outcomes stay equally likely on the 4th toss, each with probability \(\displaystyle \tfrac12\).

  4. Exercise 14

    A bag contains slips numbered from 1\displaystyle 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 12\displaystyle \frac{1}{2}. Justify.

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    NCERT’s answer
    Yes, the outcomes 'odd number', 'even number' are equally likely in the situation considered.
    Among the slips $\displaystyle 1$ to $\displaystyle 100$: \[\text{odd slips: } 1, 3, 5, \dots, 99 \quad \Rightarrow \quad n(\text{odd}) = 50 \] \[\text{even slips: } 2, 4, 6, \dots, 100 \quad \Rightarrow \quad n(\text{even}) = 50 \] \[P(\text{odd}) = \frac{50}{100} = \frac{1}{2}, \qquad P(\text{even}) = \frac{50}{100} = \frac{1}{2} \] Answer: Correct: odd and even are equally likely because $\displaystyle 50$ slips are odd and $\displaystyle 50$ are even.