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stich3 [128]
2 years ago
6

A six-faced die is rolled .

Mathematics
1 answer:
STatiana [176]2 years ago
8 0

Answer:

[A] 2/6

[B] 1/2

Step-by-step explanation:

Given:

A six-faced die is rolled .

To Find:

Find the probability of it landing on a value more than 4

Find the probability of it landing on an even number

Part A:

Knowing that a six-faced die has six number on it (1,2,3,4,5,6)

Thus, the probability of it landing on an value more than 4 is 2/6.

Because 5 and 6 is more than 4

Part B:

1,3,5 = Odd

2,4,6 = Even

Thus, the probability of it landing on an even number is 3/6 Simplify 1/2 . Which also is 50%.

<em>~Lenvy~</em>

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The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it
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Answer:

(a) 11.25 and 1.68  

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....

where,  n = number of trials (samples) taken = 15

             r = number of success

             p = probability of success which in our question is % of cases in

                  which customer problems are resolved on the same day, i.e.;75%

So, here X ~ Binom(n=15,p=0.75)

(a) Expected number of problems to be resolved today = E(X)

            E(X) = \mu = n * p = 15 * 0.75 = 11.25

    Standard deviation = \sigma = \sqrt{n*p*(1-p)} = \sqrt{15*0.75*(1-0.75)} = 1.68

(b) Probability that 10 of the problems can be resolved today = P(X = 10)

     P(X = 10) = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}

                    = 3003*0.75^{10} *0.25^{5} = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

    = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}

    = 3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4} = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

    given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)  

= \binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}

= 1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}

= 0.6865

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<h3>What are the types of probability distributions ?</h3>

Discrete and continuous probability distributions are the two main categories in which probability distributions fall. There are numerous varieties of probability distributions within each category.

  • The number of favorable outcomes of an event divided by the total number of possible outcomes of an event can be used to calculate the probability of an event.
  • The likelihood of a sure or certain event is one.
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To learn more about Probability, visit

brainly.com/question/3069881

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Answer:

Umm nine ways?

Step-by-step explanation:

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