Variance problem with probability?

Question

Three cards are drawn sequentially from a deck that contains 16 cards numbered 1 to 16 in an arbitrary order. Suppose the first card drawn is a 6.

Define the event of interest, A, as the set of all increasing 3-card sequences, i.e. A={(x1,x2,x3)|x1 < x2 < x3}, where x1,x2,x3∈{1,⋯,16}. Define event B as the set of 3-card sequence that starts with 6, i.e. B={(x1,x2,x3)|x1=6} or simply B={(6,x2,x3)}

Let Sx3=t represent the subset {(6,x2,t)|6 < x2 < t}, then |A∩B|=∑t=816∣∣Sx3=t∣∣.

Then what is |A∩B|=?

? · Accepted Answer

well, this is the way i see the problem :
concerning the notation :
 S(t) is sufficient as event description, as the dumb variable x3 is not used :
so, for example : for t >= 8, written inextenso,
S(8) = { (6,7,8) }---------------------------------1 element : t=8 and t - 7 = 1 element
S(9) =  { (6,7,9) , (6,8,9) }--------------------2 elements: t=9 and t-7 = 2 elements
etc.
S(16) = { (6,7,16) , (6,8,16) }----------------t= 16, so t - 7 = 9 elements

the classification explanation by introducing A is superfluous 
as it is renewed is S(t)

therefore the cardinal (numbers of elements) of | A n B | is :

| A n B | = ∑ (from 8 to 16 of) (t - 7)  by changing the index u = t - 7, t = u + 7 from u=1 to 16
 
 | A n B | = ∑ (from 1 to 9 of) u = 1 + 2 + ... + 9 = 9 . 10 / 2

finally

| A n B | = 45

et voilà !!

hope it' ll help !!

Randy P · Answer

"Variance problem with probability?"
No it's not. Variance is not mentioned here. Nothing here has anything to do with variance. It's just a counting problem.

= |Sx3=8| + |Sx3=9| + |Sx3=10| + ... + |Sx3=16|

Work out each of those. Add them up. For instance, here's |Sx3=11|, the number of 3 card sequences (6, x, 11). That means x can be 7, 8, 9 or 10. There are 4 of those.

Now do that for 8, 9, ..., 16.

? · Answer

!!!!

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Variance problem with probability?

3 Answers