Probability Question from Gregs Lecture

Hello,

So there was one problem where we had a set of 2 men and 2 women, call them m1 m2 w1 w2, the question was “what’s the probability of getting a man and women”. Greg did 2/4*2/3 = 4/12 which is the probability of getting a man and then a women, he then added 4/12, because you may also get a women and then a man.

In this problem however, he doesnt multiply by the various orders that may happen when choosing from these 3 distinct positive numbers, why? As in, we can get ced or cde and so on

Both questions are attached.

Thank you in advance :slight_smile:

Capture

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In the men and women question, you had 2 types of data - men and women. Here also we have 2 types of data - positive and negative numbers. In the first case, you could choose a man, and then choose a woman or choose a woman and then choose a man, right? so you were choosing either of the two types. But in the above question, you should always choose thr positive integer. Because all your selections should be positive. So, it’s like if you were asked to choose 2 men out of the first dataset. Your only choice would be to choose a man first and a man even in the second try. and you wouldn’t need any other cases, like choosing a woman first (coz that would be an invalid condition)

It’s actually better to approach the first problem like this. You can choose anyone on the first try. so 4/4. Now after choosing a person, if you’ve chosen a man, now restrict yourself to choose a woman out of the remaining 3 people, i.e. 2/3 (or vice versa). So now in just a single equation you have handled all possible cases. You can try to implement the same in the number line problem.

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Thank you for your help, that makes things clear. Have a good day!