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Greeley [361]
3 years ago
7

Imagine a deck of 500 cards numbered from 1 to 500. If all the cards are shuffled randomly and you are asked to pick three cards

, one at a time, what's the probability of each subsequent card being larger than the previous drawn card
Mathematics
1 answer:
amm18123 years ago
6 0

Answer:

P = 1 / 6

Step-by-step explanation:

Given:-

- A deck has 500 cards

- Numbered from 1-500

Find:-

you are asked to pick three cards, one at a time, what's the probability of each subsequent card being larger than the previous drawn card

Solution:-

- Suppose we draw three cards. We don't care what they actually are, irrespective of number.

- Let the three numbers be a, b, c: Such that:

                                       a < b < c.

- The total possible combinations with 3 numbered cards will be:

                                       a b c

                                       b a c

                                       a c b

                                       b a c

                                       b c a

                                       c a b

- We have 6 possibilities for 3 numbered cards. So the probability of  of each subsequent card being larger than the previous drawn card would be:

                                     P = 1 / possible combinations

                                    P = 1 / 6

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Cody and Jordan were selling Flamin' Hot Cheetos to raise money for the Tornados. Cody sold 5 bags of Cheetos for every 3 bags J
andrew11 [14]

They sold 160 and for each 5 bags sold by one 3 bags are sold by another 3+5=8.

160/8=20.

Now Since Cody Sold 5 Bags for every 3 bags for Jordan

5x20=100

3x20=60

Cody Sold 100 and Jordan Sold 60

7 0
3 years ago
Read 2 more answers
Which expression can be used to find 32% of 130?
Gemiola [76]

Answer:

b. 0.32 • 130

Step-by-step explanation:

Change the percent to decimal form

32% = .32

.32 * 130

6 0
3 years ago
The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that
salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

Part a

Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0
4 years ago
The difference between two numbers is 10 and their sum is 20. Find the biggest number.
marysya [2.9K]

Answer:

15

Step-by-step explanation:

x - y = 10, then x = 10 + y

x + y = 20

substitute for x:

10 + y + y = 20

2y = 10

y = 5

x = 10 + 5 = 15

8 0
3 years ago
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