If ten coins are tossed simultaneously, then the probability of getting at most 1 head is:
Model the 10 tosses as a binomial with \(n=10\), \(p=\tfrac{1}{2}\). "At most 1 head" means \(X = 0\) or \(X = 1\).
Step 1 — P(0 heads):
\(P(X=0) = \binom{10}{0}\left(\tfrac{1}{2}\right)^{10} = \dfrac{1}{1024}\)
Step 2 — P(1 head):
\(P(X=1) = \binom{10}{1}\left(\tfrac{1}{2}\right)^{10} = \dfrac{10}{1024}\)
Step 3 — Add:
\(P(X \le 1) = \dfrac{1 + 10}{1024} = \dfrac{11}{1024}\)
Therefore, the required probability is \(\dfrac{11}{1024}\).
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