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Evgesh-ka [11]
2 years ago
8

When dealing with the number of occurrences of an event over a specified interval of time or space and when the occurrence or no

noccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval, the appropriate probability distribution is a _____.
Mathematics
1 answer:
jeyben [28]2 years ago
7 0

Answer:

POISSON DISTRIBUTION

Step-by-step explanation:

When dealing with the number of occurrences of an event over a specified interval of time or space, the poisson distribution is often useful.

Poisson distribution is applicable if:

The probability of the occurrence of the event is the same for any two intervals of equal length.

The occurrence or nonoccurrence of the event in any interval is independent of the occurrence or nonoccurrence in any other interval.

The probability that two or more events will occur in an interval approaches zero as the interval becomes smaller.

Therefore, the appropriate probability distribution is POISSON PROBABILITY DISTRIBUTION.

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Suppose you roll a pair of honest dice. If you roll a total of 7 you win $22, if you roll a total of 11 you win $66, if you roll
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Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Number of favorable outcomes = 7

Formula for probability:

Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}

So, the possibility of getting a total of 7 = \frac{6}{36}=\frac{1}{6}

The possible ways of getting a total of 11,

{(5,6), (6,5)}

So, the probability of getting a total of 11 = \frac{2}{36} = \frac{1}{18}

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = \frac{28}{36} = \frac{7}{9}

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11

\frac{11}{3}+\frac{11}{3}-\frac{77}{9}

\frac{22}{3}-\frac{77}{9}

\frac{66-77}{9}

-\frac{11}{9}

-1.22

Therefore the expected payoff for this game is -$1.22.

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