Probability of Independent Coin Toss
The question asks for the probability of getting a head when the 6th coin is tossed.
- We are given 6 unbiased coins.
- The first 5 coins were tossed independently and resulted in heads.
- The 6th coin is now tossed independently.
Key Principle: For unbiased coins, each toss is an independent event. The outcome of previous tosses (the first 5 coins) has no influence on the outcome of the 6th toss.
Since the 6th coin is unbiased:
- The probability of getting a head ($P(H)$) is equal to the probability of getting a tail ($P(T)$).
- Mathematically, $P(H) = P(T) = \\frac{1}{2}$.
Therefore, the probability of getting a head on the 6th independent toss is $\\frac{1}{2}$.
Important Questions from Probability (Notes)
- A stick of length L is broken into two pieces at random. What is the average length of the smaller piece?
- A fair die was thrown three times and the outcome was repeatedly six. If the die is thrown again what is the probability of getting six?
A canal system is shown in the figure.
Water flows from A to B through two channels. Gates $G_1$ and $G_2$ are operated independently to regulate the flow. Probability of $G_1$ to be open is 10% while that of $G_2$ is 20%. The probability that water will flow from A to B is
- A standard fair die is rolled until some face other than 5 or 6 turns up. Let $X$ denote the face value of the last roll, and $A = \{X \text{ is even}\}$ and $B = \{X \text{ is at most 2}\}$. Then,
Consider a Markov chain having state space $S = \{1,2,3,4\}$ with transition probability matrix $P = (p_{i,j})$ given by
$$\begin{matrix} & \begin{matrix} 1 & \quad 2 & \quad 3 & \quad 4 \end{matrix} \\ \begin{matrix} 1 \\ 2 \\ 3 \\ 4 \end{matrix} & \begin{bmatrix} 1/2 & 0 & 1/2 & 0 \\ 1/4 & 1/4 & 1/4 & 1/4 \\ 1/3 & 0 & 1/3 & 1/3 \\ 1/2 & 0 & 1/2 & 0 \end{bmatrix} \end{matrix}$$
Then
