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Serggg [28]
2 years ago
15

Suppose that a red die and a green die are tossed and the numbers on the sides that face upward are observed. ​(a) what is the p

robability that the sum of the numbers is less than 4
Mathematics
1 answer:
Tems11 [23]2 years ago
5 0

First of all, let's count how many events compose our sample space: since we are rolling two dice, and each of them has six sides, the sample space has 36 possible events, given by all couples (x,y), with x,y \in \{1,2,3,4,5,6\}

Then, we observe that none of the die can give 4 or more, otherwise the sum of the two dice would be at least 5.

So, we have the following possible outcomes:

Red die gives 1, green die gives 3

Both dice give 2

Red die gives 3, green die gives 1.

These are three events out of the 36 possible events, so their probability is

\cfrac{3}{36} = \cfrac{1}{12}

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Answer:

A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.

B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.

C. The expected value of X is 6.75, and the standard deviation of X is 2.17.

Step-by-step explanation:

The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.

With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.25^{k} 0.75^{25-k}\\\\\\

A. P(x=6)

P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\

B. P(x≥10)

P(x\geq10)=1-P(x

P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.25*0.001=0.0063\\\\\\P(x=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.0625*0.0013=0.0251\\\\\\P(x=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.0156*0.0018=0.0641\\\\\\P(x=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.0039*0.0024=0.1175\\\\\\P(x=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.001*0.0032=0.1645\\\\\\P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.0002*0.0042=0.1828\\\\\\

P(x=7) = \dbinom{25}{7} p^{7}(1-p)^{18}=480700*0.000061*0.005638=0.1654\\\\\\P(x=8) = \dbinom{25}{8} p^{8}(1-p)^{17}=1081575*0.000015*0.007517=0.1241\\\\\\P(x=9) = \dbinom{25}{9} p^{9}(1-p)^{16}=2042975*0.000004*0.010023=0.0781\\\\\\

P(x\geq10)=1-(0.0008+0.0063+0.0251+0.0641+0.1175+0.1645+0.1828+0.1654+0.1241+0.0781)\\\\P(x\geq10)=1-0.9287=0.0713

C. The expected value (mean) and standard deviation of this binomial distribution can be calculated as:

E(x)=\mu=n\cdot p=25\cdot 0.25=6.25\\\\\sigma=\sqrt{np(1-p)}=\sqrt{25\cdot 0.25\cdot 0.75}=\sqrt{4.69}\approx2.17

4 0
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