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A recent school survey found that 31%31\% of Sasha's classmates speak at least two languages. If 33 of her classmates are chosen at random, what is the probability that 00 speak at least two languages? Write your answer as a decimal rounded to the nearest thousandth.____

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Q. A recent school survey found that 31%31\% of Sasha's classmates speak at least two languages. If 33 of her classmates are chosen at random, what is the probability that 00 speak at least two languages? Write your answer as a decimal rounded to the nearest thousandth.____
  1. Use Binomial Probability Formula: Use the binomial probability formula P(X=k)=C(n,k)⋅(p)k⋅(1−p)(n−k)P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}, where nn is the number of trials, kk is the number of successes, and pp is the probability of success.\newlineHere, n=3n = 3, k=0k = 0, and p=0.31p = 0.31 (probability that a classmate speaks at least two languages).
  2. Calculate C(3,0)C(3, 0): Calculate C(3,0)C(3, 0), which is the number of ways to choose 00 successes from 33 trials.\newlineC(3,0)=3!0!×(3−0)!=11×1=1C(3, 0) = \frac{3!}{0! \times (3 - 0)!} = \frac{1}{1 \times 1} = 1.
  3. Calculate (0.31)0(0.31)^0: Calculate (0.31)0(0.31)^0, which is the probability of 00 successes.\newline(0.31)0=1(0.31)^0 = 1, because any number to the power of 00 is 11.
  4. Calculate (1−0.31)(3−0)(1 - 0.31)^{(3 - 0)}: Calculate (1−0.31)(3−0)(1 - 0.31)^{(3 - 0)}, which is the probability of 33 failures.\newline(1−0.31)(3−0)=(0.69)3=0.69×0.69×0.69(1 - 0.31)^{(3 - 0)} = (0.69)^3 = 0.69 \times 0.69 \times 0.69.
  5. Multiply Values Together: Multiply all the values together to find the probability. P(X=0)=1×1×(0.69)3=(0.69)3P(X = 0) = 1 \times 1 \times (0.69)^3 = (0.69)^3.
  6. Solve (0.69)3(0.69)^3: Solve (0.69)3(0.69)^3.(0.69)3=0.69×0.69×0.69=0.328509(0.69)^3 = 0.69 \times 0.69 \times 0.69 = 0.328509.
  7. Round to Nearest Thousandth: Round the answer to the nearest thousandth.\newlineP(X=0)=0.328509P(X = 0) = 0.328509 rounded to the nearest thousandth is 0.3290.329.

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