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In a certain game, you perform three tasks. You flip a quarter, and

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In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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New post 08 Mar 2015, 20:05
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In a certain game, you perform three tasks. You flip a quarter, and success would be heads. You roll a single die, and success would be a six. You pick a card from a full playing-card deck, and success would be picking a spades card. If exactly one of these three tasks is successful, then you win the game. What is the probability of winning?

A. 1/48
B. 5/16
C. 11/12
D. 11/16
E. 23/48

Kudos for a correct solution.

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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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New post 08 Mar 2015, 22:22
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Hi Bunuel

Event 1(A) = Flipping the quarter( Probability of both winning/loosing = 1/2)
Event 2(B) = Rolling a dice( Probability of winning = 1/6 ; Probability of loosing = 1 - 1\6 = 5\6)
Event 3(C) = Drawing a card( SPADES) Probability of winning = 13/52=3/4 ; Probability of loosing = 1 - 3/4 = 1/4)

So now as above we have three events A,B & C.

Now the question says one wins when he/she exactly wins on either of the three events.
Cases in which it's a win( the highlighted green event is a win and red is loose.)
1. ABC = 1/2*5/6*3/4 = 15/48
OR
2. ABC = 1/2*1/6*3/4 = 3/48
OR
3. ABC = 1/2*5/6*1/4 = 5/48

Now as we now OR = +

P of winning = 15/48 + 3/48 + 5/48 = 23/48
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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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New post 10 Mar 2015, 09:15
1
The answer is E

Probability of just one enent will be happining


So, Probability heads=1/2

Probability number 6 =1/6

Probability picking a spades card =1/4

so, Probability win by getting heads= 1/2*5/6*3/4 = 15/48

Probability win by getting number 6 = 1/2*1/6*3/4 = 3/48

Probability win by picking a spades card =1/2*5/6*1/4=5/48

Probability wining= 15/48+3/48+5/48 =23/48
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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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New post 15 Mar 2015, 20:54
Bunuel wrote:
In a certain game, you perform three tasks. You flip a quarter, and success would be heads. You roll a single die, and success would be a six. You pick a card from a full playing-card deck, and success would be picking a spades card. If exactly one of these three tasks is successful, then you win the game. What is the probability of winning?

A. 1/48
B. 5/16
C. 11/12
D. 11/16
E. 23/48

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This is very tricky. We have to think of three cases.

Case One: success with coin, no success with die or card
P(coin = H AND die ≠ 6 AND card ≠ spade) = (1/2)*(5/6)*(3/4) = 15/48

Case Two: success with die, no success with coin or card
P(coin = T AND die = 6 AND card ≠ spade) = (1/2)*(1/6)*(3/4) = 3/48

Case Three: success with card, no success with die or coin
P(coin = T AND die ≠ 6 AND card = spade) = (1/2)*(5/6)*(1/4) = 5/48

The winning scenario could be Case One OR Case Two OR Case Three. Since these are joined by OR statements and are mutually exclusive, we simply add the probabilities.

P(win game) = 15/48 + 3/49 + 5/48 = 23/48.

Answer = (E)
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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New post 16 Mar 2015, 16:07
Event(1): Probability of getting head in a flip of coin = 1/2
Event(2): Probability of getting 6 in a roll of dice = 1/6
Event(3): Probability of getting spade in a pack of card = 1/4

Probability of winning is having exactly one successful event is:
P(success 1)* P(Fail 2)* P(Fail 3) + P(Fail 1)* P(Success 2)* P(Fail 3) + P(Fail 1)* P(Fail 2)* P(Success 3)
= 1/2*5/6*3/4 +1/2*1/6*3/4 +1/2*5/6*1/4
=15/48 + 3/48 +5/48
=23/48

E
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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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New post 16 Mar 2015, 21:18
Bunuel wrote:
Bunuel wrote:
In a certain game, you perform three tasks. You flip a quarter, and success would be heads. You roll a single die, and success would be a six. You pick a card from a full playing-card deck, and success would be picking a spades card. If exactly one of these three tasks is successful, then you win the game. What is the probability of winning?

A. 1/48
B. 5/16
C. 11/12
D. 11/16
E. 23/48

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This is very tricky. We have to think of three cases.

Case One: success with coin, no success with die or card
P(coin = H AND die ≠ 6 AND card ≠ spade) = (1/2)*(5/6)*(3/4) = 15/48

Case Two: success with die, no success with coin or card
P(coin = T AND die = 6 AND card ≠ spade) = (1/2)*(1/6)*(3/4) = 3/48

Case Three: success with card, no success with die or coin
P(coin = T AND die ≠ 6 AND card = spade) = (1/2)*(5/6)*(1/4) = 5/48

The winning scenario could be Case One OR Case Two OR Case Three. Since these are joined by OR statements and are mutually exclusive, we simply add the probabilities.

P(win game) = 15/48 + 3/49 + 5/48 = 23/48.

Answer = (E)


It should be 11/16 (1-5/16)
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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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New post 17 Mar 2015, 05:16
MyaimHarvard wrote:
Bunuel wrote:
Bunuel wrote:
In a certain game, you perform three tasks. You flip a quarter, and success would be heads. You roll a single die, and success would be a six. You pick a card from a full playing-card deck, and success would be picking a spades card. If exactly one of these three tasks is successful, then you win the game. What is the probability of winning?

A. 1/48
B. 5/16
C. 11/12
D. 11/16
E. 23/48

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This is very tricky. We have to think of three cases.

Case One: success with coin, no success with die or card
P(coin = H AND die ≠ 6 AND card ≠ spade) = (1/2)*(5/6)*(3/4) = 15/48

Case Two: success with die, no success with coin or card
P(coin = T AND die = 6 AND card ≠ spade) = (1/2)*(1/6)*(3/4) = 3/48

Case Three: success with card, no success with die or coin
P(coin = T AND die ≠ 6 AND card = spade) = (1/2)*(5/6)*(1/4) = 5/48

The winning scenario could be Case One OR Case Two OR Case Three. Since these are joined by OR statements and are mutually exclusive, we simply add the probabilities.

P(win game) = 15/48 + 3/49 + 5/48 = 23/48.

Answer = (E)


It should be 11/16 (1-5/16)


That's not a correct answer. Check the solution.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In a certain game, you perform three tasks.  [#permalink]

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New post 01 Mar 2019, 23:41
Not a realistic GMAT question, because you don't need to know about decks of cards for the test, but we have:

chance of 'success' with the coin: 1/2
chance of 'success' with the die: 1/6
chance of 'success' with the card: 1/4

There are three ways to win. One way: we might succeed with the coin *and* fail with the die *and* fail with the card. To find the probability that happens, we multiply each individual probability: (1/2)(5/6)(3/4)

But we might succeed with the die and fail with the other items: (1/2)(1/6)(3/4)

Or we might succeed with the card and fail with the other items: (1/2)(5/6)(1/4)

Adding the probabilities from those three cases gives the answer:

(1/2)(5/6)(3/4) + (1/2)(1/6)(3/4) + (1/2)(5/6)(1/4) = 23/(2)(6)(4) = 23/48
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Re: In a certain game, you perform three tasks.   [#permalink] 01 Mar 2019, 23:41
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