Question

pascals triangle is shown to the right. if three fair coins are tossed, the probability of 0 heads is $\frac{1}{8}$, the probability of 1 head is $\frac{3}{8}$, the probability of 2 heads is $\frac{3}{8}$, the probability of 3 heads is $\frac{1}{8}$. these probabilities relate to row 3 of the triangle. generalize this pattern to complete the following statement. if n fair coins are tossed, the probability of exactly x heads is the fraction whose numerator is entry number of row number in pascals triangle, and whose denominator is the sum of the entries in row number

Explanation:

Step1: Recall binomial - probability and Pascal's triangle relationship

The number of ways to get \(x\) successes (in this case, heads) in \(n\) independent Bernoulli trials (coin - tosses) is given by the binomial coefficient \(\binom{n}{x}\). In Pascal's triangle, the \(n\) - th row (starting from \(n = 0\)) contains the binomial coefficients \(\binom{n}{k}\) for \(k=0,1,\cdots,n\). The \(k\) - th entry in the \(n\) - th row of Pascal's triangle is \(\binom{n}{k}\). When we toss \(n\) fair coins, the probability of getting exactly \(x\) heads is \(P(X = x)=\frac{\binom{n}{x}}{2^{n}}\). The sum of the entries in the \(n\) - th row of Pascal's triangle is \(2^{n}\) (by the binomial theorem \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), when \(a=b = 1\), \((1 + 1)^n=\sum_{k=0}^{n}\binom{n}{k}=2^{n}\)).

Step2: Identify the positions in Pascal's triangle

If \(n\) fair coins are tossed, the probability of exactly \(x\) heads is the fraction whose numerator is entry number \(x\) of row number \(n\) in Pascal's triangle, and whose denominator is the sum of the entries in row number \(n\).

Answer:

The first blank is \(x\), the second blank is \(n\), and the third blank is \(n\).