Suppose that $90 \%$ of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?
- A$1 - \sum_{r=7}^{10} {^{10}C_r} (0.9)^r (0.1)^{10-r}$
- B$\sum_{r=0}^{6} {^{10}C_r} (0.9)^r (0.1)^{10-r}$
- C$\sum_{r=7}^{10} {^{10}C_r} (0.9)^r (0.1)^{10-r}$
- D$1 - \sum_{r=0}^{6} {^{10}C_r} (0.9)^r (0.1)^{10-r}$
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