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Senario:
Box A contains four equal-sized tickets, numbered 1, 2, 3 ,4
Box B contains three tickets of the same size, numbered 4, 5, 6
An experiment consists of selecting one ticket from the box A and then selecting one ticket from box B.

My answer:

sample space = {(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,4)(4,5),(4,6)}
events where sum is at least 7 = (1,6),(2,5),(2,6),(3,4),(3,5),(3,6),(4,4),(4,5),(4,6)}

there for probability = # of possibilities meet the constraint / # of posibilities in sample space
=9/12
=3/4


Please advice if there is a formula to get the same answer?

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#of possibilities in sample space = #tickets in A * #tickets in B (here $4\cdot 3=12$)

#of possibilities meet the constraint: Sum over thee cases of chosen tickets in Box :

for (x,6) all 4 tickets in Box A meets the conditions. For (x,5) on less -> 3, for (x,4) one more less ->2. The sum is 4+3+2=9.

Hope this Hint can help you to generalize your solution for arbitrary big boxes A and B.

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  • $\begingroup$ curious if you agree with my own generalization for consecutive integers below? I think it will be harder and require more manual checking for non-consecutive integers in the sets ('boxes'). $\endgroup$ – snulty Jul 25 '14 at 18:53
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Ok, you've accepted an answer. Here's a generalization for consecutive integers in the two sets.

$A=\{n,\ldots,n+m\}$, $\quad B=\{k,\ldots,k+l\}$, $n,k>0$ and $m,l\geq0$.

Let's say you want them to sum to the greater than the number $d \in \Bbb N$.

Suppose you know the least $a \in \Bbb N,\; 0\leq a \leq m$ s.t. there exists a $b \in \Bbb N, \; 0\leq b \leq l$ with $n+a+k+b\geq d$. We again choose the least of the available $b$'s.

So there are $l-b+1$ numbers $(\text{i.e. } k+b,\ldots, k+l)$ which also can be added to $n+a$.

There are $l-b+2$ numbers $(\text{i.e. } k+b-1,\ldots, k+l)$ which also can be added to $n+a+1$. and so on.

We sum these up, but there are two cases,

  1. $b\geq m-a$ and
  2. $m-a>b$

for 1). it's $\sum\limits_{i=0}^{m-a} (l-b+1)+i\quad $ numbers

and for 2). $\left(\sum\limits_{i=0}^{b} (l-b+1)+i\right) + (m-a-b)\cdot(l+1)$

We can evaluate these to be:

  1. $(m-a+1)\cdot(l-b+1)+\frac{1}{2}(m-a)\cdot(m-a+1)$
  2. $(b+1)\cdot(l-b+1)+\frac{1}{2}b\cdot(b+1)+(m-a-b)\cdot(l+1)$

Then the probability is just one of them divided by $\mid A\mid \cdot \mid B \mid$.

examples

your one: $A=\{1,2,3,4\}$ and $B=\{4,5,6\}$

$d=7, \; m=3, \; a=0, l=2,\; b=2, \;$

using formula 2). since $m-a=3>2=b$:

$(2+1)\cdot(2-2+1)+\frac{1}{2}2\cdot(2+1)+(3-0-2)\cdot(2+1)$ which is $3+3+3=9$.

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