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A surfboard factory just discovered a manufacturing error that causes some of the surfboards to crack easily. As a result, the factory recalled all of the surfboards sold in the last year. Each surfboard has a 75%75\% chance of having the defect.\newlineIf the factory recalled 33 surfboards, what is the probability that 00 of the boards have the defect?\newlineWrite your answer as a decimal rounded to the nearest thousandth.\newline____\newline

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Find probabilities using the binomial distributionright chevron icon

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Q. A surfboard factory just discovered a manufacturing error that causes some of the surfboards to crack easily. As a result, the factory recalled all of the surfboards sold in the last year. Each surfboard has a 75%75\% chance of having the defect.\newlineIf the factory recalled 33 surfboards, what is the probability that 00 of the boards have the defect?\newlineWrite your answer as a decimal rounded to the nearest thousandth.\newline____\newline
  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)}. Here, n=3n = 3, k=0k = 0, and p=0.75p = 0.75.
  2. Calculate C(3,0)C(3, 0): Calculate C(3,0)C(3, 0) which is 3!0!×(3−0)!\frac{3!}{0! \times (3 - 0)!}. This simplifies to 11 because any number factorial divided by itself is 11.
  3. Compute (0.75)0(0.75)^0: Compute (0.75)0(0.75)^0 which is 11, because any number to the power of 00 is 11.
  4. Calculate (1−0.75)(3−0)(1 - 0.75)^{(3 - 0)}: Calculate (1−0.75)(3−0)(1 - 0.75)^{(3 - 0)} which is (0.25)3(0.25)^3. This is 0.25×0.25×0.250.25 \times 0.25 \times 0.25.
  5. Multiply Values Together: Multiply all the values together. P(X=0)=1×1×(0.25)3P(X = 0) = 1 \times 1 \times (0.25)^3.
  6. Solve (0.25)3(0.25)^3: Solve (0.25)3(0.25)^3 which is 0.0156250.015625.
  7. Multiply Values from Step 55: Multiply the values from Step 55. P(X=0)=1×1×0.015625P(X = 0) = 1 \times 1 \times 0.015625.
  8. Final Probability: The final probability is 0.0156250.015625. Round this to the nearest thousandth.

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