Question:

The probability of losing at least 7 of 10 tosses with a fair coin is given by the sum of probabilities for losing exactly 7, 8, 9, or 10 tosses. This is a binomial probability problem where the number of trials $n=10$ and the probability of losing a toss (success) $p=0.5$. The probability $P(X=k)$ for $k$ losses in $n$ tosses is $\binom{n}{k} p^k (1-p)^{n-k}$. Since $p=0.5$, $1-p=0.5$, so $p^k (1-p)^{n-k} = (0.5)^k (0.5)^{10-k} = (0.5)^{10} = \left(\frac{1}{2}\right)^{10}$. Thus, the probability of losing at least 7 tosses is: $$P(X \geq 7) = \sum_{k=7}^{10} \binom{10}{k} \left(\dfrac{1}{2}\right)^{10}$$ $$P(X \geq 7) = \left(\dfrac{1}{2}\right)^{10} \left[\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}\right]$$ First, calculate the binomial coefficients: $$\binom{10}{7} = \dfrac{10!}{7!3!} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$ $$\binom{10}{8} = \dfrac{10!}{8!2!} = \dfrac{10 \times 9}{2 \times 1} = 45$$ $$\binom{10}{9} = \dfrac{10!}{9!1!} = 10$$ $$\binom{10}{10} = \dfrac{10!}{10!0!} = 1$$ Now, substitute these values back into the equation: $$P(X \geq 7) = \dfrac{1}{2^{10}} [120 + 45 + 10 + 1]$$ $$P(X \geq 7) = \dfrac{1}{1024} [176]$$ $$P(X \geq 7) = \dfrac{176}{1024}$$ To simplify the fraction, divide both the numerator and denominator by their greatest common divisor. $$P(X \geq 7) = \dfrac{176 \div 16}{1024 \div 16} = \dfrac{11}{64}$$