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Jet001 [13]
3 years ago
7

If you flip a coin and roll a 6-sided die, what is the probability that you will flip a tails and roll at least a 2?

Mathematics
1 answer:
Finger [1]3 years ago
4 0

Step-by-step explanation:

Flipping a tails and rolling at least a tail are 2 independent events: they don't affect each other. So, to get the probability of both happening, we just need to multiply the probability of one by the probability of the other. he probability of rolling at least a 2 is 5/6.

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2. Is the best choice
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The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it
brilliants [131]

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

3 0
3 years ago
If there are 75 people at a movie and 36% of them are men, how many men are at the movie?
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75 * .36 = 27 men

****************************************************************


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