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choli [55]
3 years ago
11

Because gambling is a big​ business, calculating the odds of a gambler winning or losing in every game is crucial to the financi

al forecasting for a casino. Consider a slot machine that has three wheels that spin independently. Each has 13 equally likely​ symbols: 5 ​bars, 4 ​lemons, 3 ​cherries, and a bell. If you play​ once, what is the probability that you will​ get:
Mathematics
1 answer:
IRISSAK [1]3 years ago
5 0

Answer:

a) 0.1165

b) 0.0983

c) 0.000455

d) 0.787

e) 0.767

Step-by-step explanation:

5 ​bars, 4 ​lemons, 3 ​cherries, and a bell

Total = 5+4+3+1 = 13

The probability of getting a bar on a slot, P(Ba) = 5/13 = 0.385

A lemon, P(L) = 4/13 = 0.308

A cherry, P(C) = 3/13 = 0.231

A bell, P(Be) = 1/13 = 0.0769

a) Probability of getting 3 lemons = (4/13) × (4/13) × (4/13) = 256/2197 = 0.1165

b) Probability of getting no fruit symbol

On each slot, there are 4+3 = 7 fruit symbols.

Probability of getting a fruit symbol On a slot = 7/13

Probability of not getting a fruit symbol = 1 - (7/13) = 6/13 = 0.462

Probability of not getting a fruit symbol On the three slots = 0.462 × 0.462 × 0.462 = 0.0983

c) Probability of getting 3 bells, the jackpot = (1/13) × (1/13) × (1/13) = 1/2197 = 0.000455

d) Probability of not getting a bell on the 3 slots

Probability of not getting a bell on one slot = 1 - (1/13) = 12/13 = 0.923

Probability of not getting a bell on the 3 slots = (12/13) × (12/13) × (12/13) = 1728/2197 = 0.787

e) Probability of at least one bar is a sum of probabilities

Note that Probability of getting a bar = 5/13 and probability of not getting a bar = 8/13

1) Probability of getting 1 bar and other stuff on the 2 other slots (this can happen in 3 different orders) = 3 × (5/13)×(8/13)×(8/13) = 960/2197 = 0.437

2) Probability of getting 2 bars and other stuff on the remaining slot (this can also occur in 3 different orders) = 3 × (5/13)×(5/13)×(8/13) = 600/2197 = 0.273

3) Probability of getting 3 bars on the slots machine = (5/13) × (5/13) × (5/13) = 125/2197 = 0.0569

Probability of at least one bar = 0.437 + 0.273 + 0.0569 = 0.7669 = 0.767

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According to the National Vital Statistics, full-term babies' birth weights are Normally distributed with a mean of 7.5 pounds a
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Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

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