Suppose we have a group of 10 light users and 10 heavy users. what is the probability that exactly 3 of the 20 users are hiv positive?

The answer is 30%.

Step-by-step explanation:  Given that there is a group of 10 light users and 10 heavy users. We are to find the probability that exactly 3 of the 20 users are HIV positive.

We have the following four possible cases -

(i) All 3 are light users.

(ii) 1 is a light user and 2 are heavy users.

(iii) 2 are light users and 1 is a heavy user.

(iv) All 3 are heavy user.

Since there is a 45% chance of a light user to be HIV positive and 55% chance of a heavy user to be HIV positive, so the required probability is given by

Thus, the probability is 30.

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

1) is right

step-by-step explanation: