MHB What's the probability of selecting John and then James as disciples of Jesus?

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The probability of selecting John and then James as disciples of Jesus is calculated based on the total number of disciples, which is 12. The probability of choosing John first is 1/12, and after selecting John, the probability of choosing James next is 1/11. Therefore, the combined probability of selecting John and then James is 1/(12*11), resulting in 1/132. The discussion emphasizes the importance of understanding the problem-solving process rather than just receiving answers. Engaging with the material step-by-step is crucial for deeper comprehension.
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Jesus had 12 disciples. What's the probability of selecting John and then James?

What is the set up?
 
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Basically, you keep asking same question over and over again! Are you actually learning anything?

If you have n objects, the probability of choosing anyone of them is 1/n. If you have n objects, once you have chosen one, without replacement, there are n- 1 objects left. The probability of choosing anyone of those n- 1 objects is 1/(n-1). When the probability of event A is p and the probability of event B is q, the probability of "event A and event B" is pq.

Here n= 12. The probability of choosing "John" first is 1/12. The probability of choosing "James" next is 1/11. The probability of choosing "John" and then "James" is 1/(11*12)= 1/132.
 
Thank you very much.
 
Hey RTCNC,

I do agree that this question is similar to the one you asked yesterday, so for me that only highlights the fact that you should work with us step-by-step to truly understand these problems. If someone just solves it for you, you won't be able to easily transfer that to other problems. Why don't you post what you've tried and your thoughts on it? Do you agree with what I'm saying conceptually?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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