Problem 1
Question
One hundred identical coins, each with probability \(p\) of showing up heads, are tossed. If \(0
Step-by-Step Solution
Verified Answer
The value of \( p \) is \( \frac{50}{101} \) (Option C).
1Step 1: Understand the Probability Condition
The problem states that the probability of getting exactly 50 heads is equal to the probability of getting exactly 51 heads. Mathematically, this can be expressed as \( P(X=50) = P(X=51) \), where \( X \) is the random variable representing the number of heads.
2Step 2: Define Binomial Probabilities
The number of heads, \( X \), follows a binomial distribution with parameters \( n = 100 \) and \( p \). The probability of getting exactly \( k \) heads is given by \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \).
3Step 3: Set Up the Probability Equation
Using the binomial formula, set up the equation for the probabilities: \[ \binom{100}{50} p^{50} (1-p)^{50} = \binom{100}{51} p^{51} (1-p)^{49}. \]
4Step 4: Simplify the Equation
Divide both sides by \( \binom{100}{50} p^{50} (1-p)^{49} \) to simplify: \( (1-p) = \frac{100 - 50 + 1}{51} p \). This simplifies to \( 1-p = \frac{51}{51} p \).
5Step 5: Solve for p
Equate and solve the simplified equation: \( 1 - p = \frac{50}{51}p \). Rearrange to get \( 1 = \frac{101}{51} p \). Solve for \( p \) to find \( p = \frac{50}{101} \).
Key Concepts
ProbabilityRandom VariablesBinomial Coefficient
Probability
Probability measures the chance of an event occurring and is a fundamental concept in statistics. It ranges from 0 to 1, where 0 indicates impossibility and 1 indicates certainty.
In the context of the binomial distribution, probability plays a pivotal role in determining the likelihood of a specific outcome. For example, when tossing a coin, each side has a probability, or chance, of coming up based on fairness and randomness of the tosses.
When dealing with multiple events, such as flipping a hundred coins like in our problem, probability helps calculate the chance of getting a particular number of heads. In our exercise, we are specifically interested in the situation where the probability of getting 50 heads equals the probability of getting 51 heads among the 100 coins tossed.
Probability uses mathematical expressions and formulas to quantify these chances, which then allows us to make predictions and decisions based on those calculations.
Random Variables
A random variable is a variable that assumes numerical values based on the outcomes of a random phenomenon. It is a fundamental concept in probability theory and statistics. Two types of random variables exist: **discrete** and **continuous**. In the coin-tossing problem, the random variable is discrete because it represents the number of heads (which can only take on integer values between 0 and 100).In our scenario, the random variable \(X\) represents the number of heads obtained from flipping 100 coins. By defining \(X\), we can determine probabilities for different outcomes using the properties of the binomial distribution.Understanding random variables is key to applying probability distributions and analyzing the likelihood of various outcomes in real-world situations.
Binomial Coefficient
The binomial coefficient is a crucial element in the binomial distribution formula. It is denoted as \( \binom{n}{k} \). This notation defines the number of ways to choose \( k \) successes (like heads) from \( n \) experiments (like coin tosses).The binomial coefficient can be calculated using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \( n! \) represents the factorial of \(n\), which is the product of all positive integers up to \(n\).In our exercise with 100 coin tosses, we use binomial coefficients to find the probability of getting exactly 50 or exactly 51 heads. Here, \( \binom{100}{50} \) and \( \binom{100}{51} \) are instrumental in setting up the equation that allows us to solve for the value of \( p \). This connection between combinatorics and probability theory helps simplify complex probability calculations.
Other exercises in this chapter
Problem 2
If \(A\) and \(B\) are two events such that \(P(A \cup B) \geq \frac{3}{4}\) and \(\frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8}\), then (A) \(P(A)+P(B) \leq \f
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A point is selected at random from the interior of a circle. The probability that the point is closer to the centre than the boundary of the circle is (A) \(\fr
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From a box containing 20 tickets of value 1 to 20 , four tickets are drawn one by one. After each draw, the ticket is replaced. The proability that the largest
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