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MartyMurray
At an end-of-year party for the employees of a certain company, Martha has been given an opportunity to win as many as two prizes. She must draw, at random, exactly one numbered card from each of two boxes—each of which, in turn, contains exactly six cards, numbered from 1 to 6. If the sum of the two numbers is equal to 8, then Martha will win a tablet computer. And, if at least one of the two numbers is a 2, then Martha will win a smartphone.

Select the probability that Martha will win a Tablet computer, and select the probability that Martha will win Both prizes. Make only two selections, one in each column.


Since the answer for first column is the probability that Martha will win just the tablet and the answer for the second column is the probability that she will win both prizes, we will use constraints that we discover in finding the answer for the first column in finding the answer for the second column. So, let's find the answer for the first column first.

For Martha to win a tablet computer, the sum of the numbers on the two cards must be 8.

The total number of ways she can draw the two cards is 6 x 6 = 36.

The ways she can draw cards whose numbers sum to 8 are the following:

2 and 6

3 and 5

4 and 4

5 and 3

6 and 2

Since there are 5 such pairs, the probability that Martha will win a tablet computer is 5/36.

Then, of the 5 ways she can win a tablet computer, 2 include a 2. So, she has 2 ways to win both a tablet computer and a smartphone. Thus, the probability that she'll win both prizes is 2/36 or 1/18.

Correct Answer
5/36, 1/18

Could you explain why have not you included 4 and 4 pair twice?
 
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Could you explain why have not you included 4 and 4 pair twice?
Interesting question. Here's why.

­She's just drawing one card from each box. Order doesn't make any difference.

In other words, 5 and 3 and 3 and 5, for example, are different only because she has drawn different cards from the boxes in the two cases, but there's only one way to draw two 4s. You just draw a 4 from each box.­­
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At an end-of-year party for the employees of a certain company, Martha has been given an opportunity to win as many as two prizes. She must draw, at random, exactly one numbered card from each of two boxes—each of which, in turn, contains exactly six cards, numbered from 1 to 6. If the sum of the two numbers is equal to 8, then Martha will win a tablet computer. And, if at least one of the two numbers is a 2, then Martha will win a smartphone.

Select the probability that Martha will win a Tablet computer, and select the probability that Martha will win Both prizes. Make only two selections, one in each column.
To calculate the ­probability of winning a tablet computer, we need to find the number of ways of getting a sum of 8.
8 = 2+6 = 3+5 = 4+4

P(2+6) = 1/6 * 1/6 * 2 = 2/36 (we multiply by 2 because she could draw the card 2 from either box and the 6 from the other)

Same will be the probability of obtaining 3+5 but for 4+4 we will not multiply by 2 (because the numbers are same, 4 each)

Hence Total Probability = 2/36 + 2/36 + 1/36 = 5/36

Now probability of winning both will be less than 5/36 (the probability of winning just the tablet).
The only value less than 5/36 is 1/18.

ANSWER: 5/36, 1/18

If you need to calculate the probability of both, notice that she needs (2+6) or (3+5) or (4+4) to get a tablet and she will also win the phone only if she gets (2+6). Hence the probability of winning both is simply the probbaility of getting (2+6) which is 2/36.­

Check the discussion on Probability here:
https://youtu.be/0BCqnD2r-kY
­
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The phone alone would have a probability of 1/3, right? 6*2 would account for all combinations with a 2 from both boxes, divided by 36= 12/36=1/3.­
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I don't think so.. The passage info tells us that Martha wins a smartphone if at least 1 of the dice is 2..

This means she wins if she gets a combo of (2, x) or (x, 2)

I took this as p(win) = 1 - loss scenario
Loss scenario = 5/6 (possibility of getting a number that is not 2 on the first card) * 5/6 (possibility of getting a number that is not 2 on the second card)

Therefore scenario of win = 1 - 5/6 * 5/6 = 11/36..

You can also think of this as win scenarios = 2,1 2,2 2,3 2,4 2,5 2,6 when 2 on the first dice
and 1,2 2,2 3,2 4,2 5,2 6,2 when 2 on the second dice..

This is equal to 11 scenarios (12 -1 [since 2,2 is counted in both scenarios]) / 36 possible scenarios = 11/36
SergejK
The phone alone would have a probability of 1/3, right? 6*2 would account for all combinations with a 2 from both boxes, divided by 36= 12/36=1/3.­
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"She must draw, at random, exactly one numbered card from each of two boxes" was anyone confused by the phrase exactly one numbered card ? Is this the right structure to form the sentence ? i wasted time trying to understand if she will be rewarded if she draws 2 ones
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Yes, the sentence is grammatically correct... But it could be clearer.

If this slowed you down, mentally rephrase the sentence in simpler terms to anchor what is happening whenever you see these word combinations.

"she draws 2 cards, one from each box" for example.

RahulGowdaK
"She must draw, at random, exactly one numbered card from each of two boxes" was anyone confused by the phrase exactly one numbered card ? Is this the right structure to form the sentence ? i wasted time trying to understand if she will be rewarded if she draws 2 ones
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