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Kitty [74]
3 years ago
11

A slot machine at a hotel is configured so that there is a 1/1200 probability of winning the jackpot on any individual trial. If

a guest plays the slot machine 6 times, find the probability of exactly 2 jackpots. If a guest told the hotel manager that she had hit two jackpots in 6 plays of the slot machine, would the manager be surprised?
Mathematics
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer:

The probability is approximately one in a million (0.001% ).

It's a very little probability, so the manager would be really surprised.

Step-by-step explanation:

We have for this question to calculate with a binomial random variable, with n=6 and p=1/1200=0.00083 .

We have to calculate the probability of getting 2 hits in the sample of size 6.

P(x=2)=\binom{6}{2}p^2(1-p)^4=15*(1/1200)^2*(1199/1200)^4\\\\P(x=2)=15* 0.000000694*0.996670831= 0.00001038

The probability is approximately one in a million (0.001% ).

It's a very little probability, so the manager would be really surprised.

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You can factor the 32 out of the sum:

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We can also change the index as follows

\displaystyle 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m}

Now, we have a theorem that states that the series

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\displaystyle \sum_{m=1}^\infty a^m = \dfrac{1}{1-a}

This is your case, because you have

|a|=\dfrac{1}{2}

which implies that your series converges, and the value is

\displaystyle 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m} = 32 \cdot \dfrac{1}{1-\frac{1}{2}} = 32\cdot 2 = 64

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