A biased coin with the probability of getting head equal to \(\frac{1}{4}\) is tossed five times. What is the probability of getting tail in all the first four tosses followed by head ?
Understanding the Biased Coin Probability Problem
This problem involves calculating the probability of a specific sequence of outcomes when tossing a biased coin multiple times. A biased coin means that the probability of getting a head is not equal to the probability of getting a tail. We are given the probability of getting a head and need to find the probability of a particular sequence of five tosses.
Defining Probabilities for the Biased Coin
We are told that the probability of getting a head (H) with this biased coin is \(\frac{1}{4}\). In probability theory, the sum of the probabilities of all possible outcomes of an event must equal 1. For a coin toss, the only two outcomes are head (H) or tail (T).
- Probability of Head, \(P(H) = \frac{1}{4}\)
- Probability of Tail, \(P(T)\)
So, the probability of getting a tail is \(P(T) = 1 - P(H)\).
Let's calculate the probability of getting a tail: $$P(T) = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$$ Thus, the probability of getting a tail in a single toss is \(\frac{3}{4}\).
Probability of a Specific Sequence of Independent Events
The coin is tossed five times. Each coin toss is an independent event, meaning the outcome of one toss does not affect the outcome of any other toss. We are interested in the probability of a specific sequence of outcomes: getting a tail in the first four tosses followed by a head in the fifth toss.
The desired sequence is: Tail, Tail, Tail, Tail, Head (TTTTH).
To find the probability of a sequence of independent events, we multiply the probabilities of each individual event in the sequence.
Probability of the sequence TTTTH = \(P(T) \times P(T) \times P(T) \times P(T) \times P(H)\).
Calculating the Probability of the TTTTH Sequence
Now, we substitute the probabilities we found:
$$P(\text{TTTTH}) = \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{1}{4}\right)$$Let's perform the multiplication:
$$P(\text{TTTTH}) = \frac{3 \times 3 \times 3 \times 3}{4 \times 4 \times 4 \times 4} \times \frac{1}{4}$$ $$P(\text{TTTTH}) = \frac{3^4}{4^4} \times \frac{1}{4}$$ $$P(\text{TTTTH}) = \frac{81}{256} \times \frac{1}{4}$$ $$P(\text{TTTTH}) = \frac{81 \times 1}{256 \times 4}$$ $$P(\text{TTTTH}) = \frac{81}{1024}$$So, the probability of getting tail in all the first four tosses followed by head is \(\frac{81}{1024}\).
| Toss Number | Desired Outcome | Probability |
|---|---|---|
| 1st Toss | Tail (T) | \(\frac{3}{4}\) |
| 2nd Toss | Tail (T) | \(\frac{3}{4}\) |
| 3rd Toss | Tail (T) | \(\frac{3}{4}\) |
| 4th Toss | Tail (T) | \(\frac{3}{4}\) |
| 5th Toss | Head (H) | \(\frac{1}{4}\) |
The probability of the sequence TTTTH is the product of these individual probabilities: \(\left(\frac{3}{4}\right)^4 \times \left(\frac{1}{4}\right)^1 = \frac{81}{256} \times \frac{1}{4} = \frac{81}{1024}\).
Revision Table: Biased Coin Probability
| Concept | Explanation | How Applied Here |
|---|---|---|
| Biased Coin | Probability of outcomes (H/T) is not equal. | \(P(H) = \frac{1}{4}\), so \(P(T) = \frac{3}{4}\). |
| Independent Events | Outcome of one event does not affect others. | Each coin toss is independent. |
| Probability of Sequence | For independent events, multiply individual probabilities. | \(P(TTTTH) = P(T) \times P(T) \times P(T) \times P(T) \times P(H)\). |
Additional Information: Understanding Probability
Probability is a measure of the likelihood of an event occurring. It is a number between 0 and 1, where 0 means the event is impossible and 1 means the event is certain.
- Sample Space: The set of all possible outcomes of an experiment. For a single coin toss, the sample space is {H, T}. For five tosses, the sample space is much larger, containing \(2^5 = 32\) possible sequences.
- Event: A specific outcome or set of outcomes from the sample space. In this problem, the specific event is the sequence TTTTH.
- Calculating Probability: For simple events with equally likely outcomes, probability is often calculated as (\(\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\)). However, for biased coins or specific sequences of independent events, we use the probabilities of the individual outcomes as demonstrated above.
Understanding independent events is crucial for solving many probability problems, especially those involving repeated trials like coin tosses or dice rolls.
Important Questions from Probability of Random Experiments
A, B, C and D are mutually exclusive and exhaustive events.
If 2P(A) = 3P(B) = 4P(C) = 5P(D), then what is 77P(A) equal to ?
A fair coin is tossed 6 times. What is the probability of getting a result in the 6t h toss which is different from those obtained in the first five tosses ?
Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is
Three dice are thrown. What is the probability that each face shows only multiples of 3 ?
A natural number n is chosen from the first 50 natural numbers. What is the probability that \(n+\frac{50}{n}<50 \) ?
