Question:

Let $X$ be the number of tosses lost out of 10. Since the coin is fair, the probability of losing a toss is $p = 1/2$. The number of tosses is $n = 10$. The probability of losing $k$ tosses out of $n$ is given by the binomial probability formula: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ In this case, $p = 1/2$ and $1-p = 1/2$, so $p^k (1-p)^{n-k} = (1/2)^k (1/2)^{n-k} = (1/2)^n$. Thus, $$P(X=k) = \binom{10}{k} \left(\frac{1}{2}\right)^{10}$$ Step 1: Calculate the probability of losing at least 7 of 10 tosses. This is $P(X \geq 7)$, which is the sum of probabilities for $X=7, 8, 9, 10$. $$P(X \geq 7) = \sum_{k=7}^{10} P(X=k) = \sum_{k=7}^{10} \binom{10}{k} \left(\frac{1}{2}\right)^{10}$$ $$P(X \geq 7) = \left(\frac{1}{2}\right)^{10} \left[ \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} \right]$$ Step 2: Calculate the binomial coefficients and sum them. $$\binom{10}{7} = \frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$$ $$\binom{10}{8} = \frac{10!}{8!2!} = \frac{10 \times 9}{2 \times 1} = 45$$ $$\binom{10}{9} = \frac{10!}{9!1!} = 10$$ $$\binom{10}{10} = 1$$ Sum of coefficients: $120 + 45 + 10 + 1 = 176$. Step 3: Calculate the final probability. Since $(1/2)^{10} = 1/1024$: $$P(X \geq 7) = \frac{1}{1024} \times 176 = \frac{176}{1024}$$ To simplify the fraction, divide both numerator and denominator by their greatest common divisor. $$176 = 16 \times 11$$ $$1024 = 16 \times 64$$ $$P(X \geq 7) = \frac{16 \times 11}{16 \times 64} = \frac{11}{64}$$