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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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08 Mar 2015, 19:01
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00:00

Difficulty:

85% (hard)

Question Stats:

54% (02:10) correct 46% (01:50) wrong based on 271 sessions

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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 any of these task are 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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In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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08 Mar 2015, 20:51
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1
Hi Bunuel

According to me dr are two ways to solve the question

First lets get the overview

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.

1st method(This is lengthy method)
Possible 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
OR
2.ABC = 1/2*1/6*3/4
OR
3.ABC = 1/2*5/6*1/4
OR
4.ABC = 1/2*1/6*3/4
OR
5.ABC = 1/2*5/6*1/4
OR
6.ABC = 1/2*5/6*1/4
OR
7.ABC
= 1/2*1/6*1/4

As we now OR means add
Adding up all of them we'll get the probability of winning ie 11/16

2nd Method (For those whose who like short methods)

dr's a concept that Probability of not happening = 1 - Probability of happening

This is what we gonna use here

No let's calculate Probability of loosing in all the three events

P of loosing in A = 1 - 1/2 = 1/2
P of loosing in B = 1 - 1/6 = 5/6
P of loosing in C = 1 - 1/4 = 3/4

Therefore P of loosing the game = (P of loosing in A) * (P of loosing in B) * (P of loosing in C) = 1/2*5/6*3/4 = 5/16

Now P of winning = 1 - 5/16 = 11/16
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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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08 Mar 2015, 21:11
1
Hi All,

Either by accident or by design, the answer choices are written in such a way that you can get to the correct answer with a minimal amount of math (and a bit of logic).

We're told that there are 3 ways to win a "game":
1) Flip a coin and get 'heads'
2) Roll a die and get a "6"
3) Pick a random card from a standard deck of cards and get a "spade"

When performing all 3 tasks, if you succeed in ANY of them, then you 'win.' The answers are all fractions (and we can take advantage of those specific answers).

First, the probability of winning on the coin flip is 1/2; when you consider that there are 2 additional ways to win (if you 'lose' at the coin flip), then the probability of 'winning' MUST be greater than 1/2. Eliminate Answers A, B and E (they're all LESS than 1/2).

With the two remaining options (11/12 and 11/16), there's enough of a disparity that we can work logically from the given information. The probability of 'winning' on the die roll is 1/6 and the probability of 'winning' on the card is 1/4, so the probability of winning each of those events is relatively low. 11/12 is over 90%, which is MUCH TOO HIGH relative to the individual probabilities involved. As such, we can eliminate C.

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

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10 Mar 2015, 08:02
2

Probability of enent not happening is equal to (1 - Probability of event happening)

Probability NO number 6 =5/6

Probability NO picking a spades card =3/4

so, Probability NO winning = 1/2*5/6*3/4 = 15/48=5/16

Probability NO winning = 1 - 5/16 = 11/16 .
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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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15 Mar 2015, 19:51
1
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 any of these task are 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:

In this scenario, winning combinations would include success on any one task as well as any combination of two or three successes. In other words, there are several cases that constitute the winning combinations. By contrast, the only way to lose the game would be unsuccessful at all three tasks. Let’s use the complement rule.

P(lose game) = P(quarter = T AND dice ≠ 6 AND card ≠ spades)
= (1/2)*(5/6)*(3/4) = 5/16

P(win game) = 1 – P(lose game) = 1 – (5/16) = 11/16

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

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05 Jul 2015, 04:32
Hi, here's another approach (the shortest way is the original version from Magoosh)
1/2 + 1/2*1/6 + 1/2*5/6*1/4 = 33/48 = 11/16 (D)
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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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28 Oct 2016, 06:06
My approach was the same...find the probability of not winning...
IT IS IMPERATIVE THAT WE READ THE QUESTION ATTENTIVELY!!!
If any of these task are successful, then you win the game.

the question basically asks: what is the probability of winning AT LEAST ONCE???
to get to the answer, it's faster to find the probability of not winning at all:

not win 1 = 1/2
not win 2 = 5/6
not win 3 = 3/4

1/2 * 5/6 * 3/4 = 5/16
probability of winning is thus 1 - 5/16 = 11/16
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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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07 Nov 2018, 18:02
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 any of these task are 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

We can let A = event of getting heads when flipping the quarter, B = event of getting a six when rolling the die and C = event of getting a spades card, and use the following formula:

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)

P(A or B or C) = 1/2 + 1/6 + 1/4 - (1/2 x 1/6) - (1/2 x 1/4) - (1/4 x 1/6) + (1/2 x 1/6 x 1/4)

P(A or B or C) = 11/12 - 1/12 - 1/8 - 1/24 + 1/48

P(A or B or C) = 11/16

Alternate Solution:

We notice that P(success) + P(failure) = 1; therefore, P(success) = 1 - P(failure). Let’s find P(failure).

The only way we fail in this game is if we get tails from the quarter flip AND not get a six from the die roll AND not get a spade from the card draw. Therefore,

P(failure) = 1/2 x 5/6 x 3/4 = 15/48 = 5/16

Thus, P(success) = 1 - P(failure) = 1 - 5/16 = 11/16

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

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10 Nov 2018, 19:18
In this question why can't we multiply= p(occurring head) * p(occurring 6) * p( picking spade).
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Re: In a certain game, you perform three tasks. You flip a quarter, and  [#permalink]

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11 Nov 2018, 12:16
Hi Roma123,

The prompt tells us that if you complete ANY of the 3 tasks, then you win the game. The calculation that you are referring to is the probability of winning ALL of the tasks. While that is one way to win the game, it is NOT all of the possible ways to win it.

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Re: In a certain game, you perform three tasks. You flip a quarter, and   [#permalink] 11 Nov 2018, 12:16
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