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.Tom is throwing darts at a target. In his last 30 throws, Tom has hit the target 20 times. You want to know the estimated probability that exactly one out of the next three throws does not hit the target.

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MathPhys

Step-by-step explanation:

In his last 30 throws, Tom hit the target 20 times and missed 10 times.

p = probability he misses the target = ⅓

q = probability he hits the target = ⅔

Using binomial probability:

P = nCr (p)^r (q)^(n-r)

Given n = 3, r = 1, p = ⅓, and q = ⅔:

P = ₃C₁ (⅓)¹ (⅔)³⁻¹

P = (3) (⅓) (⅔)²

P = 4/9

There is a 4/9 probability that he misses exactly 1 of his next 3 throws.

Lanuel

By applying binomial probability, the estimated probability that exactly one (1) out of the next three (3) throws doesn't hit the target is 4/9.

How to determine the estimated probability?

In order to determine the estimated probability that exactly one (1) out of the next three (3) throws doesn't hit the target, we would apply binomial probability equation:

[tex]P =\; ^nC_r (p)^r (q)^{(n-r)}[/tex]

Based on the information given, we can deduce the following points:

  • Probability that Tom misses the target (p) = 10/30 = ⅓
  • Probability that Tom hits the target (q) = 20/30 = ⅔
  • Number of next throws (n) = 3.
  • Number of throws taken at a time (r) = 1.

Substituting the given parameters into the equation, we have;

P = ³C₁ × (⅓)¹ × (⅔)³⁻¹

P = 3 × (⅓)¹ × (⅔)²

P = 3 × ⅓ × (⅔)²

P = 1 × 4/9

P = 4/9.

Read more on probability here: https://brainly.com/question/25870256

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