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, ...).
Let $P_{odd}$ be the probability that the number of tosses required is odd.
Consider the outcome of the first toss:
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} $The probability that the number of required tosses is odd is $2/3$. This corresponds to Option C.
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)
For a distribution with mean, median, mode and standard deviation 25, 24, 26 and 5 respectively, Karl Pearson's coefficient of skewness equals to:
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: