There are 5 students in a classroom, 3 from Group A and 2 from Group B. 3 apples were given to 3 out of the 5 students randomly. Calculate the probability that: i) All 3 Group A students received the apples, ii) At least 2 out of 3 Group A students received the apples.
what are you struggling with?
I don't know where to start
So let's try to think of the problem differently. Think of being given an apple as being CHOSEN to eat an apple. So then, we're choosing 3 people to eat apples :). So what's the probability that the first person chosen is from group A?
3/5
and the second person?
2/4
wait so its 3/5 times 2/4 times 1/3 = 1/10
yes :)
I'm stuck with part ii) though
mmhmm.... so are you familiar with permutations and combinations?
um no
so does at least 2 mean 2 or 3, or does it mean just 3?
2 or 3
Because we just did 3, we can just calculate the probability for 2 and add on to the probability for 3. How do we calculate the probability of two? Well, generally we'd use a formula (combinations), but let's not worry about it... here are the possibilities for choices where 2 are A AAB ABA BAA Did I miss any? how many are there?
3 combinations
Alright, so the interesting thing is that we can calculate the probability of AAB, ABA, and BAA separately, but you will actually find they turn out to be the same, because the denominator will be the same and only the order of the numbers will change. Since any of these situations count as at least 2 apples being given to A, we need to calculate the probability of them all happening. In this case though, because they are the same, we can just do case AAB and then multiply that probability by 3. And what would case AAB be?
3/5 x 2/4 x 2/3 = 1/5
So what's the probability of 2 apples being given to group A? and using that, the probability of at least 2 apples being given to group A?
3/5, then add to 1/10 to get 7/10
yup!
Ah, I get it. Thank you :)
No problem :)
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