Twelve envelopes are on a table. Four contain $15 each, thre

Question

Twelve envelopes are on a table. Four contain $15 each, three contain $50 each, and five contain $80 each. If four envelopes are chosen at random, what is the probability that the total amount of money in the four envelopes exceeds $230?

(A) $\frac{7}{33}$
(B) $\frac{2}{9}$
(C) $\frac{8}{33}$
(D) $\frac{3}{11}$
(E) $\frac{1}{3}$

(A)  \frac{7}{33} 
(B)  \frac{2}{9} 
(C)  \frac{8}{33} 
(D)  \frac{3}{11} 
(E)  \frac{1}{3}

stne · Answer

Number of ways we can get more than  230 .

All four  80  ->  5_C_4 = 5

Three  80  and one 15 ->  5_C_3 * 4_C_1 = 10 *4 = 40

Three 80 and one 50 ->  5_C_3 * 3_C_1 =10 *3 = 30

Two 80 and two 50 ->  5_C_2 * 3_C_2 = 10*3= 30

Total favourable cases =  5+40+30+30 = 105

Total cases :  12_C_4 = 495

Prob  = \frac {105}{495}= \frac {7}{33}

Ans A

Hope it helped.

alamin8 · Answer

Yes, I did the same way! Thank you for clarification.

malvikasachdev124 · Answer

why is 5C3*9C1 not correct?

egmat · Answer

Hi malvikasachdev124,

Your instinct is right that "3 eighties plus 1 more" is a genuine case here. The trouble is that  9C1  is quietly doing two things you did not intend: it counts some groups more than once, and at the same time it leaves one whole case out.

The leftover 9 still contains eighties

There are five  $80  envelopes. Once  5C3  takes  3  of them, the leftover pile of  9  is: four  $15 , three  $50 , and  two more $80s .

So when  9C1  picks one of those two eighties, you have not built a "3 eighties + 1 other" group at all. You have built a  four eighties  group.

Why that group gets counted 4 times

Label the eighties E1 to E5, and say the four you end up holding are E1, E2, E3, E4. Your method can build that exact same set in four different ways:

- three = E1, E2, E3, extra = E4
- three = E1, E2, E4, extra = E3
- three = E1, E3, E4, extra = E2
- three = E2, E3, E4, extra = E1

Same four envelopes on the table, counted  4  times over. stne's  5C4 = 5  line counts each such group exactly  once , which is what you actually want.

Split your 90 apart and you can see it

5C3 * 9C1 = 10 * 9 = 90 . Break the  9  up by what the extra envelope turns out to be:

- Extra is a  $15  or a  $50 , so  7  choices:  10 * 7 = 70 . This part is completely correct, and it is exactly stne's  40 + 30 = 70 .
- Extra is an  $80 , so  2  choices:  10 * 2 = 20 . This should have been  5 , so you are  15  too high.

One case never shows up at all

5C3  forces at least three eighties into every group. So your count can never reach  two  eighties +  two  fifties =  160 + 100 = 260 , which does clear  $230  and is worth  30  groups ( 5C2 * 3C2 ).

Put both corrections together:  90 - 15 + 30 = 105 , which is stne's total.  Therefore105/495 = 7/33 .

The rule worth carrying forward

"Pick a fixed block, then pick freely from the rest" is only safe when the leftover pile has  none  of that same item type left in it. The moment the rest still contains the same kind of item, one final group can be assembled several different ways and your count inflates. Splitting by how many come from each dollar group, the way stne did, makes that impossible.

Answer: A

malvikasachdev124 · Answer

Thank You So Much!!!!!!!!!!!

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