All Exams Test series for 1 year @ ₹349 only
Question

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 ?

The correct answer is \(\frac{1}{32}\)

Probability of a Different Outcome in Coin 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.

Understanding the Problem: Fair Coin 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.

Total Possible Outcomes for 6 Tosses

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 = $2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.

Identifying Favorable Outcomes

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:

  • If the first five tosses include both Heads and Tails (e.g., HHTHT), can the 6th toss be different from *all* of them? No. If the 6th toss is Head, it is the same as the H outcomes in the first five. If the 6th toss is Tail, it is the same as the T outcomes in the first five. Therefore, if the first five tosses are mixed, the 6th toss cannot be different from *all* of them.
  • For the 6th toss to be different from *all* of the first five, the first five tosses must *all* be the same.

There are two possible scenarios where the first five tosses are all the same:

  1. All first five tosses are Heads (HHHHH). For the 6th toss to be different, it must be a Tail (T). This gives the sequence: HHHHHT.
  2. All first five tosses are Tails (TTTTT). For the 6th toss to be different, it must be a Head (H). This gives the sequence: TTTTTH.

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.

Calculating the Probability

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:

  • Number of Favorable Outcomes = 2
  • Total Number of Possible Outcomes = 64

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 = $\frac{2 \div 2}{64 \div 2} = \frac{1}{32}$.

Conclusion on Probability

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)

Revision Table: Coin Toss Probability Concepts

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 $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$

Additional Information: Understanding Coin Toss Problems

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).

Was this answer helpful?

Important Questions from Probability of Random Experiments

  1. 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 ?

  2. Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is

  3. 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 ? 

  4. Three dice are thrown. What is the probability that each face shows only multiples of 3 ?

  5. A natural number n is chosen from the first 50 natural numbers. What is the probability that \(n+\frac{50}{n}<50 \) ?

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App