• Expected Number of Binomial Outcomes:
• Number of tosses per set: $n = 10$
• Probability of head on a single toss: $p = 0.5$, tails: $q = 0.5$
• Probability of getting exactly 7 heads and 3 tails in a set of 10 tosses:

$$P(X = 7) = \binom{10}{7} p^7 q^3 = \binom{10}{3} (0.5)^{10} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times \frac{1}{1024} = \frac{120}{1024} = \frac{15}{128} \approx 0.1171875$$

• Total number of sets conducted: $N = 100$
• The expected number of sets showing 7 heads is:

$$E = N \times P(X = 7) = 100 \times 0.1171875 = 11.72 \approx 12$$

• Conclusion: 12 cases are expected (Option 1).

Correct option: (1) 12