George has a fair 5-sided spinner with sides labelled 1, 2, 3, 4, 5. He spins the spinner and notes the number on the side on which the spinner lands.
Find the probability that it takes fewer than 7 spins for George to obtain a 5.
Solution
The probability of landing on a 5 in one spin is \(P(5) = 0.2\) since the spinner is fair and has 5 sides.
We need to find the probability that it takes fewer than 7 spins to get a 5. This is the complement of not getting a 5 in all 6 spins.
The probability of not getting a 5 in one spin is \(1 - 0.2 = 0.8\).
The probability of not getting a 5 in 6 spins is \(0.8^6\).
Therefore, the probability of getting at least one 5 in fewer than 7 spins is:
\(P(X < 7) = 1 - 0.8^6\)
Calculating this gives:
\(1 - 0.8^6 = 1 - 0.262144 = 0.737856\)
Thus, the probability that it takes fewer than 7 spins for George to obtain a 5 is approximately 0.73856.
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