Problem 29
Question
If a coin be tossed \(n\) times, then probability that the head comes odd number of times is (a) \(1 / 2\) (c) \(1 / 2^{n-1}\) (d) None of these
Step-by-Step Solution
Verified Answer
The probability is \(1/2\), which is option (a).
1Step 1: Understanding the Problem
We need to find the probability that when a coin is tossed \(n\) times, heads appear an odd number of times. A coin toss has two outcomes: heads or tails. We are concerned with cases where the number of heads is odd.
2Step 2: Analyzing Outcomes Using Binomial Distribution
Each coin toss is an independent event, with heads or tails as outcomes, each having probability \(\frac{1}{2}\). The number of heads in \(n\) tosses follows a binomial distribution \(B(n, \frac{1}{2})\). The probability of getting \(k\) heads is given by \(P(X = k) = \binom{n}{k} \left(\frac{1}{2}\right)^n\).
3Step 3: Identifying Odd Number of Heads
For the probability of odd heads, we are interested in the sum of probabilities where \(k\) is odd. That is, \(P(X = 1) + P(X = 3) + ... + P(X = n)\) if \(n\) is odd, or up to \(P(X = n-1)\) if \(n\) is even.
4Step 4: Using Symmetry in Binomial Distribution
From the properties of binomial distribution, the outcomes where heads appear are symmetric. Hence, the number of cases with odd heads is equal to the number of cases with even heads. Thus, the probability of getting either is \(\frac{1}{2}\).
5Step 5: Conclusion and Answer Verification
Since the probability that a coin shows heads an odd number of times equals the probability it shows heads an even number of times due to symmetry, both are \(\frac{1}{2}\). Thus, the correct choice is \(\frac{1}{2}\), which matches option (a).
Key Concepts
Probability TheoryOdd Number of OutcomesCoin Toss Experiments
Probability Theory
Probability theory is a mathematical framework used to analyze the likelihood of various outcomes in random phenomena. It helps us predict how often events will occur. For a coin toss, which can end in heads or tails, probability theory provides a way to systematically calculate the chances of each of these outcomes.
- **Sample Space**: In probability theory, the set of all possible outcomes is called the sample space. For a single coin toss, this is {Heads, Tails}.
- **Event**: An event is any subset of the sample space. For example, getting heads is an event.
- **Probability**: The probability of an event is calculated by dividing the number of successful outcomes (favorable to the event) by the total number of outcomes in the sample space. For a fair coin, the probability of getting heads is \( \frac{1}{2} \).
Odd Number of Outcomes
An odd number of outcomes refers to the particular count of events that results in an odd number, such as 1, 3, or 5 occurrences. In the context of coin tosses, it means counting how many times heads appear in an odd number.
To compute this probability, consider:
To compute this probability, consider:
- Outcome Set: For 3 tosses, the possible number of times heads can occur ranges from 0 to 3. Here, the odd numbers are 1 and 3.
- Summation of Probabilities: We calculate the individual probabilities of getting these odd counts and add them together. For an experiment with multiple flips, the calculations involve summing probabilities from the binomial distribution formula.
Coin Toss Experiments
Coin toss experiments are classic examples in probability theory used to illustrate basic principles in a straightforward manner. In each toss:
- Outcomes: There are two outcomes: getting heads or tails, each with a probability of \( \frac{1}{2} \).
- Independence: Each toss is independent, meaning the result of one toss does not affect another.
- **Example:** With 2 tosses, there are 4 possible results (HH, HT, TH, TT). With 3 tosses, there are 8 possible outcomes.
- **Binomial Distribution:** This model helps us understand probabilities of combinations of results, like getting exactly 2 heads in 3 tosses.
Other exercises in this chapter
Problem 28
A bag contains 4 white, 5 red and 6 black balls. If 2 balls are drawn at random, then the probability that one of them is white is (a) \(44 / 105\) (b) \(11 / 1
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One card is drawn from the pack of cards. It is now replace in the pack and a card is again drawn. If it is done six times, then the probability of coming 2 car
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Six positive and 8 negative numbers are given. If 4 numbers are chosen and multiplied, then the probability of getting a positive product is (a) \(15 / 1001\) (
View solution Problem 30
Two events \(A\) and \(B\) are such that \(P(A)=1 / 4\) \(P(B / A)=1 / 2, P(A / B)=1 / 4\), then \(P(\bar{A} / \bar{B})\) is equal to (a) \(1 / 4\) (b) \(3 / 4\
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