The integers 1 through 5 are written on each of five cards. The cards

Question

The integers 1 through 5 are written on each of five cards. The cards are shuffled and one card is drawn at random. That card is then replaced, the cards are shuffled again and another card is drawn at random. This procedure is repeated one more time (for a total of three times). What is the probability that the sum of the numbers on the three cards drawn was 14 or 15?

A. 1/125

B. 2/125

C. 4/125

D. 1/25

E. 2/25

A. 1/125

B. 2/125

C. 4/125

D. 1/25

E. 2/25

AntrikshR · Answer

Sum will be 14 or 15 iff the cards drawn are (5,5,4) or (5,5,5)
No.of ways we can get (5,5,4): 3!/2! = 3
No.of ways we can get (5,5,5):1
Probability of getting a 5 is 1/5
Probability of getting a 4 is 1/5
=> Probability of getting sum as 14/15 = 3*(1/5)^3 +1*(1/5)^3
=> 4*(1/5)^3 or 4/125 (C)

OR
Total number of cases: 5^3
Total number of favorable outcomes: 4 (555,554,545,455)
P(14/15) :4/125.

Fighter1095 · Answer

probability = (favourable outcomes)/total number of outcomes

15 or 14 is possible when (555) (455)(545)(554)

p = (1+((3!)/(2!))/(5*5*5)

p = 4/125

IanStewart · Answer

There are 5^3 different selections (in order) that can be made. Of those selections, exactly four of them produce a sum of 14 or 15: 555, 455, 545, 554. So the answer is 4/5^3 = 4/125.

MathRevolution · Answer

Total numbers: 5

Probability of each card:  \frac{1}{5}

For sum of 15, all cards have to be '5' in three rounds. Hence,  \frac{1 }{ 5} * \frac{1}{5} * \frac{1}{5}  =  \frac{1}{125}

For a sum of 14, two cards have to be '5' and one card has to be '4' in the overall three rounds. Hence,  \frac{1 }{ 5} * \frac{1}{5} * \frac{1}{5}  =  \frac{1}{125} . But these three cards can have  \frac{3! }{ 2!}  arrangement = 3 ways. Therefore, 3 *  \frac{1}{125}  =  \frac{3 }{125}

Overall for 14 or 15:  \frac{3 }{ 125}  +  \frac{1 }{ 125}  =  \frac{4 }{ 125}

Answer C

flippedeclipse · Answer

There are two possible outcomes that work: (5,5,4) and (5,5,5). Order matters here, and there is replacement.

Calculating the (5,5,4) outcome first:

\frac{1}{5}*\frac{1}{5}*\frac{1}{5}=\frac{1}{125}

There are  \frac{3!}{2!}=3  ways to arrange the (5,5,4) arrangement, so we can multiply the above by 3 to get  \frac{3}{125} .

For the (5,5,5) arrangement, it results in  \frac{1}{5}*\frac{1}{5}*\frac{1}{5}=\frac{1}{125} , and there is only one way to arrange this.

Since these are mutually exclusive outcomes, we add these probabilities together.

\frac{1}{125}+\frac{3}{125}=\frac{4}{125}

luisdicampo · Answer

Deconstructing the Question  
We draw a card from {1,2,3,4,5}  three times with replacement .
So outcomes are  ordered triples  (because the 1st/2nd/3rd draw matter).
We want  P(	ext{sum}=14 	ext{ or } 15) .

Step-by-step  
 Total outcomes: 
Each draw has 5 possibilities, and draws are independent:
 5^3 = 125

Sum = 15: 
The maximum possible sum is  5+5+5=15 , so only:
 (5,5,5) 
That is  1  outcome.

Sum = 14: 
To get 14 (one less than 15), we must have two 5s and one 4:
 5+5+4=14 
The 4 can be in any of the 3 positions, giving  3  outcomes:
 (4,5,5), (5,4,5), (5,5,4)

Favorable outcomes: 
 1 + 3 = 4

So,
 P=\frac{4}{125}

Answer:     (C) 4/125

The integers 1 through 5 are written on each of five cards. The cards

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GMAT Problem Solving (PS) Questions

The integers 1 through 5 are written on each of five cards. The cards

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