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Question

A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

The correct answer is
$2/3$

Problem Analysis: Probability of Odd Tosses for First Head

The question asks for the probability that the number of tosses required to get the first head from a fair coin is an odd number (1, 3, 5, ...).

  • A fair coin implies the probability of getting a head (H) is $p = 1/2$, and the probability of getting a tail (T) is $q = 1 - p = 1/2$.
  • This scenario follows a geometric distribution, where the probability of the first success (head) occurring on the $k$-th trial is $P(X=k) = q^{k-1}p$.

Calculating the Probability

Let $P_{odd}$ be the probability that the number of tosses required is odd.

Consider the outcome of the first toss:

  • Case 1: First toss is a Head (H). The probability of this is $p = 1/2$. The number of tosses is 1, which is odd. So, this case contributes $p$ to $P_{odd}$.
  • Case 2: First toss is a Tail (T). The probability of this is $q = 1/2$. We have used one toss, and now we need the *remaining* tosses to result in the first head appearing on an *even* number of additional tosses. This means the total number of tosses will be $1 + (\text{even number})$, which is odd. The process essentially restarts after the first tail, but we require the subsequent process to finish in an even number of steps for the total to be odd. The probability of needing an even number of additional tosses is $1 - P_{odd}$. This case contributes $q \times (1 - P_{odd})$ to $P_{odd}$.

Combining these cases:

$ P_{odd} = P(\text{First toss H}) \times 1 + P(\text{First toss T}) \times P(\text{Remaining tosses needed are even}) $ $ P_{odd} = p \times 1 + q \times (1 - P_{odd}) $

Substitute the values $p = 1/2$ and $q = 1/2$:

$ P_{odd} = \frac{1}{2} + \frac{1}{2} \times (1 - P_{odd}) $

Now, solve for $P_{odd}$:

$ P_{odd} = \frac{1}{2} + \frac{1}{2} - \frac{1}{2} P_{odd} $ $ P_{odd} = 1 - \frac{1}{2} P_{odd} $ $ P_{odd} + \frac{1}{2} P_{odd} = 1 $ $ \frac{3}{2} P_{odd} = 1 $ $ P_{odd} = \frac{1}{3/2} $ $ P_{odd} = \frac{2}{3} $

Conclusion

The probability that the number of required tosses is odd is $2/3$. This corresponds to Option C.

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Important Questions from Basics of Probability

  1. Suppose a 5-bit message is transmitted from a source to a destination through a noisy channel. The probability that a bit of the message gets flipped during transmission is 0.01. Flipping of each bit is independent of one another. The probability that the message is delivered error-free to the destination is ______ . (rounded off to three decimal places)

  2. A quadratic polynomial $(x - \alpha)(x - \beta)$ over complex numbers is said to be square invariant if $(x - \alpha)(x - \beta) = (x - \alpha^2)(x - \beta^2)$. Suppose from the set of all square invariant quadratic polynomials we choose one at random.

    The probability that the roots of the chosen polynomial are equal is ____________ (rounded off to one decimal place)
  3. If, out of 50 fish of a pond, 12 had no ectoparasite and remaining had varying numbers of parasites on them, then how many of fish had at least one parasite ?
  4. For a distribution with mean, median, mode and standard deviation 25, 24, 26 and 5 respectively, Karl Pearson's coefficient of skewness equals to:

  5. Suppose two distributions have the same mean, same standard deviation and are equally skewed, but the first distribution is more peaked than the
    other. This feature of the frequency distributions in which they are different are called:

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