The probability of getting a “head” in a single toss of a biased coin is 0.3. The coin is tossed repeatedly till a “head” is obtained. If the tosses are independent, then the probability of getting "head" for the first time in the fifth toss is ___________
The question asks for the probability of a specific sequence of events when tossing a biased coin multiple times. The coin has a known probability of landing heads, and the goal is to find the chance of achieving the first head only on the fifth toss.
For the first head to appear on the fifth toss, the sequence must be: Tail, Tail, Tail, Tail, Head (T, T, T, T, H).
Since the tosses are independent, we multiply the probabilities of each individual outcome:
Probability = P(T) $\times$ P(T) $\times$ P(T) $\times$ P(T) $\times$ P(H)
Probability = $(P(T))^4 \times P(H)$
Substitute the values:
Probability = $(0.7)^4 \times (0.3)$
Calculate $(0.7)^4$:
$(0.7)^4 = 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401$
Now, multiply by P(H):
Probability = $0.2401 \times 0.3 = 0.07203$
The calculated probability is 0.07203. This value falls between 0.07 and 0.08, matching the provided answer range.
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: