Answer:
Step-by-step explanation:
So 25% of all students knowing how to read means if you pick a random student there is a 25% chance of them being able to read. if you picked 2 it would be .25*.25 = .0625 or 6.25% both know how to read. If you picked 10 it would be .25^10 probability that all 10 know how to read.
You want to know the probability of 2 or less (it says fewer than 3 so that's 2 or less) of the 10 being able to read. I think there is a formula, but I don't know it so I'm going to brute force it.
I am going to find the chance of one knowing how to read and two knowing then add them.
If we labeled the students A,B,C,D,E,F,G,H,I,J there are 10 different variations where just 1 can read, so we can find one and then multiply by 10. In short for one we just multiply the probabilities, so .75^9 * .25, then we multiply by 10 to represent all 10. so in total, the odds of one new student being able to read is 10*.25*.75^9
Now the chance fo exactly two knowing how to read. First, how many variations are there. It is a combination since student A and B knowing is the same as B and A. so 10C2 or 10 choose 2. If you don't know how to calculate combinations let me know. 10C2 = 45. So now if we find the probability of one possibility of two students knowing how to read we can multiply it by 45.
Now to find the probability of any two knowing, we again multiply all the probabilities. .25^2 * .75^8 then multiply by 45 so 45 * .25^2 * .75^8
Now we know the probability of exactly one knowing how to read and exactly two, so we just add them to get the probability of 1 or 2 knowing.
10*.25*.75^9 + 45 * .25^2 * .75^8
It doesn't look like any of those are the answer so i have to ask if you have the right values, otherwise I may be wrong.
