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2 years ago

P is the probability of a certain event. Which of the following indicates that P is certain to happen?

Mathematics

2 answers:

podryga [215]2 years ago

8 0

Note for probability, if it will never happen the P = 0. If it must happen P = 1.

So certain to happen:

C. P=1

8090 [49]2 years ago

8 0

The probability of the occurrence of an event ranges only from 0 to 1. Thus, eliminating the fourth choice, P = 100. If an event is not bound to happen, P = 0. If there are circumstances that may hinder it to happen, the probability is somewhere between the extremes. Lastly, if it certain to happen, the probability is 1. Therefore, the answer is letter C.

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A collection of 30 gems, all of which are identical in appearance, are supposedto be genuine diamonds, but actually contain 8 wo

levacccp [35]

Answer:

Step-by-step explanation:

From the given information:

There are 30 collections of gems, of which 8 are worthless;

Thus, the number of the genuine diamonds = 30 - 8 = 22.

Let X = random variable;

X consider the value as 0 (for 2 worthless stone selection),

X = 1200(1 worthless stone & 1 genuine stone)

X = 2400 (2 genuine stones selected)

However, the numbers of ways of selecting and chosen Gems can be estimated as:

$(^n_r) = (^{30}_2) \\ \\ \implies \dfrac{30!}{2!(30-2)!} \\ \\ \implies \dfrac{30!}{2!(28)!} \\ \\ \implies \dfrac{30*29*28!}{2!(28)!} \\ \\ \implies \dfrac{30*29}{2*1} \\ \\ \implies 435$

Thus;

$Pr (X = 0) = \dfrac{(^8_2)}{435}$

$Pr (X = 0) = \dfrac{\dfrac{8!}{2!(8-2)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8!}{2!(6)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8*7*6!}{2!(6)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8*7}{2*1}}{435} \\ \\ Pr (X = 0) = 0.0644$

$P(X =1200) = \dfrac{(^{8}_{1})(^{22}_{1})}{435}$

$P(X =1200) = \dfrac{ \dfrac{8!}{1!(8-1)!}) ( \dfrac{22!}{1!(22-1)!}) }{435}$

$P(X =1200) = \dfrac{ (8) ( 22) }{435}$

$P(X =1200) =0.4046$

$Pr (X = 2400) = \dfrac{(^{22}_2)}{435}$

$Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(22-2)!}}{435} \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(20)!}}{435} \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22*21*20!}{2!(20)!}}{435} \\ \\ Pr (X =2400) = \dfrac{\dfrac{22*21}{2*1}}{435} \\ \\ Pr (X = 2400) = 0.5310$

To find E(X):

E(X) = (0 × 0.0644) + (1200 × 0.4046) + (2400 × 0.5310)

E(X) = 0 + 485.52 + 1274.4

E(X) = 1759.92

8 0

2 years ago

\dfrac k{2}+\dfrac12=3 <br> 2<br> k<br> <br> + <br> 2<br> 1<br> <br> =3

konstantin123 [22]

it's so hard to find it

Step-by-step explanation:

I didn't know

3 0

2 years ago

Read 2 more answers

How do you calculate the minimised dimensions: length width and height of a rectangular prism with the given volume of 10m^3?

Elan Coil [88]

The length, width and height of the rectangular prism is ∛10

<h3>How to determine the dimensions?</h3>

The volume is given as:

Volume, V = 10m^3

For a minimised volume, we have:

Length = Width = Height

This means that:

V = L^3

So, we have:

L^3 = 10

Take the cube root of both sides

L = ∛10

Hence, the length, width and height of the rectangular prism is ∛10