Answer: the chance you will pull a yellow it 2/6 and the chance you pull a red is 3/6
Step-by-step explanation: add all them up and you get six and then you put the number by six
The value of y would be zero in that problem
You have to guess the first 5 digits.
You can calcualte the number of possible numbers that exist with 5 digits.
The first digit cannot be 0, it can be any from 1 to 9, so those are 9 diferent digits.
For the second, third, fourth and fifth digits there are 10 diferent posibilities for each: 0 through 9.
Then, there are 9*10*10*10*10 = 90,000 different possible social security numbers once the last four digits are known.
And the probability to get the correct social security number of the person is 1 /90,000 = 0,000011.
Answer:
The probability that a person has HIV given that the test negative is 0.0033.
Step-by-step explanation:
Denote the events as follows:
<em>X</em> = a person in Uganda had HIV
<em>Y</em> = a person in Uganda was tested positive for HIV.
The information provided is:
![P(Y^{c}\cap X)=\frac{4}{1517}\\](https://tex.z-dn.net/?f=P%28Y%5E%7Bc%7D%5Ccap%20X%29%3D%5Cfrac%7B4%7D%7B1517%7D%5C%5C)
![P(Y\cap X)=\frac{166}{1517}\\](https://tex.z-dn.net/?f=P%28Y%5Ccap%20X%29%3D%5Cfrac%7B166%7D%7B1517%7D%5C%5C)
![P(Y\cap X^{c})=\frac{129}{1517}\\](https://tex.z-dn.net/?f=P%28Y%5Ccap%20X%5E%7Bc%7D%29%3D%5Cfrac%7B129%7D%7B1517%7D%5C%5C)
![P(Y^{c}\cap X^{c})=\frac{1218}{1517}\\](https://tex.z-dn.net/?f=P%28Y%5E%7Bc%7D%5Ccap%20X%5E%7Bc%7D%29%3D%5Cfrac%7B1218%7D%7B1517%7D%5C%5C)
The probability that a person has HIV given that he/she was tested negative is:
![P(X|Y^{c})=\frac{P(Y^{c}\cap X)}{P(Y^{c})}](https://tex.z-dn.net/?f=P%28X%7CY%5E%7Bc%7D%29%3D%5Cfrac%7BP%28Y%5E%7Bc%7D%5Ccap%20X%29%7D%7BP%28Y%5E%7Bc%7D%29%7D)
Compute the probability of a person not having HIV as follows:
![P(Y^{c})=P(Y^{c}\cap X)+P(Y^{c}\cap X^{c})=\frac{4}{1517}+\frac{1218}{1517} =\frac{4+1218}{1517} =\frac{1222}{1517}](https://tex.z-dn.net/?f=P%28Y%5E%7Bc%7D%29%3DP%28Y%5E%7Bc%7D%5Ccap%20X%29%2BP%28Y%5E%7Bc%7D%5Ccap%20X%5E%7Bc%7D%29%3D%5Cfrac%7B4%7D%7B1517%7D%2B%5Cfrac%7B1218%7D%7B1517%7D%20%20%3D%5Cfrac%7B4%2B1218%7D%7B1517%7D%20%3D%5Cfrac%7B1222%7D%7B1517%7D)
Compute the value of
as follows:
![P(X|Y^{c})=\frac{P(Y^{c}\cap X)}{P(Y^{c})}=\frac{4}{1517}\times\frac{1517}{1222}=0.0033](https://tex.z-dn.net/?f=P%28X%7CY%5E%7Bc%7D%29%3D%5Cfrac%7BP%28Y%5E%7Bc%7D%5Ccap%20X%29%7D%7BP%28Y%5E%7Bc%7D%29%7D%3D%5Cfrac%7B4%7D%7B1517%7D%5Ctimes%5Cfrac%7B1517%7D%7B1222%7D%3D0.0033)
Thus, the probability that a person has HIV given that he/she was tested negative is 0.0033.
Answer:
Step-by-step explanation:
Your answer
Money left for gift c is = 5 19/20
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