Question

A Las Vegas handicapper can correctly predict the winning professional football team 70% of the time. The probability that she is wrong in her next prediction is 30%. For the same handicapper, find the probability that she is correct in each of her next two predictions.

Answer 1

Answer:

0.49 or 49%

Step-by-step explanation:

The probability of the handicapper correctly predicting a single game, P(C), is 0.7, therefore, the probability of that same handicapper being correct in each of her next two predictions, P(2C) is:

$P(2C) = P(C)^2\\P(2C) = 0.7^2\\P(2C) = 0.49$

The probability of her being correct in each of her next two predictions is 0.49 or 49%