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Probability theory predicts that there is 77.6% chance of a particular soccer player making 2 penalty shots in a row. If the soccer player taking 2 penalty shots is simulated 2500 times, in about how many simulations would you expect at least 1 missed shot?

Respuesta :

nobillionaireNobley
P(at least 1 missed shot) = 100 - 77.6 = 22.4%

Number of times to expect al least 1 missed shot = 0.224 x 2500 = 560 times.

Answer:

In about 560 simulations we would expect at least 1 missed shot.

Step-by-step explanation:

Probability theory predicts that there is 77.6% chance of a particular soccer player making 2 penalty shots in a row.

⇒ Probability of not missing a shot = 0.776

So, Probability of missing at least one shot = 1 - 0.776

                                                                      = 0.224

So, Chance of missing at least one shot = 22.4%

A shot is simulated 2500 times ⇒ n = 2500

Required simulations in which we would expect one shot missed = Probability of missing at least one shot × Number of simulations

⇒ Required simulations = 0.224 × 2500

                                          = 560

Hence, in about 560 simulations we would expect at least 1 missed shot.

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