1) A certain disease has an incidence rate of 0.4%. If the false negative rate is 7% and the false positive rate is 5%, compute the probability that a person who tests positive actually has the disease.

2 From a group of 8 people, you randomly select 3 of them.

What is the probability that they are the 3 oldest people in the group?

Give your answer as a fraction

3 You pick 6 digits (0-9) at random without replacement, and write them in the order picked.

What is the probability that you have written the first 6 digits of your phone number? Assume there are no repeats of digits in your phone number.

Give your answer as a fraction.

4 In a lottery game, a player picks six numbers from 1 to 27. If the player matches all six numbers, they win 40,000 dollars. Otherwise, they lose $1.

What is the expected value of this game?

 

(1) The probability that a person who tests positive actually has the disease is 0.65126

 

(2) The probability that they are the 3 oldest people in the group is 1 / 56

 

(3) The probability that you have written the first 6 digits of your phone number is 1 / 151200

 

(4) The expected value of this game is -0.86

 

 

 

 

Please see the given below explanation, i hope that it will clear all your doubt.

If you like the answer then please mark it as helpful. Please

 

Step-by-step explanation

According to the question,

 

(1) Using Bayes theorem

 

P(disease / positive) = P(disease) . P(positive / disease) / P(disease) . P(positive / disease) + P(no disease) . P(positive / no disease)

 

= (0.04) * (0.93) / (0.04) * (0.93) + (0.996) * (0.02)

 

= 0.0372 / 0.0372 + 0.01992

 

= 0.0372 / 0.05712

 

= 0.65126

 

0.65126 is the probability that a person who test positive actually has the disease.

 

 

(2) Given,

 

From a group of 8 people, you randomly select 3 of them

 

Total number of ways = ⁸C₃

 

= 8! / (8 - 3)! 3!

 

= 8! / 5! 3!

 

= 56 ways

 

Number of ways of selecting 3 oldest people in the group = 1 ways

 

Probability = 1 / 56

 

 

(3) The number of ways to pick an exact order of any 6 different numbers from 10 = ¹⁰P₆ = 10! / 4!

 

= 10 * 9 * 8 * 7 * 6 * 5 = 151200 ways

 

which means the first one can be any of 10 numbers

 

The second one can be any of 9 numbers

 

The 3rd one can be any of 8 numbers and so on

 

Now the 6th one can be any of 5 numbers

 

Thus there are 10 * 9 * 8 * 7 * 6 * 5 = 151200 permutations

 

Thus the required probability is 1 / 151200

 

 

(4) Probability of winning = 1 / ²⁷C₆

 

= 1 / 296010

 

Expected value = (1 / 296010) * 40,000 + (1 - 1 / 296010) * (-1)

 

= -0.86