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 ?
This question asks for the probability of a specific outcome when a fair coin is tossed multiple times. We need to find the chance that the result of the 6th toss is different from the results obtained in the first five tosses.
A fair coin has two equally likely outcomes when tossed: Heads (H) or Tails (T). The probability of getting a Head is \(\frac{1}{2}\), and the probability of getting a Tail is \(\frac{1}{2}\). Each coin toss is an independent event, meaning the outcome of one toss does not affect the outcome of any other toss.
When a coin is tossed 6 times, the total number of possible sequences of outcomes can be calculated. Since there are 2 outcomes for each toss and there are 6 tosses, the total number of possible outcomes is \(2^6\).
Total outcomes = \(2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64\).
The condition for a favorable outcome is that the result of the 6th toss is different from the results obtained in the first five tosses. Let's analyze this condition:
There are two possible scenarios where the first five tosses are all the same:
These are the only two possible sequences of 6 tosses where the 6th toss is different from every one of the first five tosses.
So, the number of favorable outcomes is 2.
The probability of an event is calculated as:
Probability = \(\frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}\)
Using the numbers we found:
Probability = \(\frac{2}{64}\)
This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 2.
Probability = \(\frac{2 \div 2}{64 \div 2} = \frac{1}{32}\).
The probability of getting a result in the 6th toss which is different from those obtained in the first five tosses is \(\frac{1}{32}\).
| Event | Description | Number of Outcomes |
|---|---|---|
| Total Outcomes | All possible sequences for 6 coin tosses | \(2^6 = 64\) |
| Favorable Outcomes | 6th toss differs from all of the first five (first five must be identical) | 2 (HHHHHT, TTTTTH) |
| Concept | Explanation |
|---|---|
| Fair Coin | Equal probability for Head and Tail (\(\frac{1}{2}\)) |
| Independent Events | Outcome of one toss does not affect others |
| Sample Space | Set of all possible outcomes (size \(2^n\) for \(n\) tosses) |
| Favorable Outcomes | Outcomes that meet the specific condition |
| Probability Formula | \(\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}\) |
Coin toss probability problems often involve understanding independent events and calculating the size of the sample space using powers of 2. When conditions link multiple tosses, carefully analyze how those conditions restrict the possible outcomes. For example, questions might ask for the probability of getting a specific number of heads, alternating results, or streaks of heads or tails. Each specific condition will define a different subset of the total outcomes as favorable.
In this particular problem, the phrasing "different from those obtained in the first five tosses" was key. It required the 6th toss to be different from *every* outcome seen in the first five. This constraint is much stronger than simply being different from the *last* toss or different from the *majority* of the first five. The only way for the 6th outcome to be different from every outcome in the first five is if the first five outcomes were all identical (either all Heads or all Tails).
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