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

Question

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.

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

iamdp · Accepted Answer

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.  A  BC  = 1/2*5/6*3/4
OR
2. A  B  C  = 1/2*1/6*3/4
OR
3. AB  C  = 1/2*5/6*1/4
OR
4. AB  C  = 1/2*1/6*3/4
OR
5. A  B  C  = 1/2*5/6*1/4
OR
6. A  BC  = 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

EMPOWERgmatRichC · Answer

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.

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

23a2012 · Answer

The answer is D

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

So, Probability NO heads=1/2

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 .

Bunuel · Answer

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

Answer = (D)

BrainLab · Answer

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)

mvictor · Answer

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

ScottTargetTestPrep · Answer

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

Answer: D

Roma123 · Answer

In this question why can't we multiply= p(occurring head) * p(occurring 6) * p( picking spade). 
please clear my doubts.

EMPOWERgmatRichC · Answer

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.

GMAT assassins aren't born, they're made,
Rich

nishaadnak · Answer

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)+(3/48)+(5/48) = 23/48​​.

Regor60 · Answer

What's missing from this calculation is the fact that you could be successful on any two/fail on the third, as well as being successful on all three.

Nothing in the question dictates that the game stop after the first successful event.

Posted from my mobile device

Ngjosworks · Answer

Posted from my mobile device

ThatDudeKnows · Answer

Plenty of great solutions posted above, so I won't repeat those. Just want to give a little tip if you decide to go the 1-(ProbabilityOfLosing) path...which is what I'd do. The trap there is taking the time to calculate the probability of losing and then forgetting to subtract from 1. Ugh. You knew what to do, did the right calculations, and then simply forgot the last step. Soooo annoying. And if you did that on a practice question or on a mock test, you'd probably just chalk it up to a careless mistake and tell yourself to be more careful next time...and then you'd make a similar careless mistake on some other question in the future. It's not simply a careless mistake. The test-writers know that you feel the pressure of the clock counting down in the corner of the screen and they set up problems that require that one little extra step at the end knowing that there's a chance you'll just forget to do that last step.

So, here's a trick to avoid falling into their little trap (or at least to do it less frequently). As soon as you spot that you're going to do some math and that there's going to be one last little step at the end, flip your computer mouse over. That way, when you go to click your answer and grab for the mouse, the fact that it's upside down will serve as your reminder to make sure you completed the last step.

Lodz697 · Answer

Why can't we do:

\frac{1}{2} * 1 * 1 + \frac{1}{2 }* \frac{1}{6} * 1 + \frac{1}{2} * \frac{5}{6 }* \frac{1}{4}  (?)

Considering that once we win the first round it is irrelevant what are the dice and cards results...  Bunuel

Bunuel · Answer

= 1/2*1/6*1/4 +

It should be;

(coin win)(die loss)(card loss) + (coin loss)(die win)(card loss) + (coin loss)(die loss)(card win) +

+ (coin win)(die win)(card loss) + (coin loss)(die loss)(card win) + (coin loss)(die win)(card win) +

+(coin win)(die win)(card win) =

= (1/2*5/6*3/4 + 1/2*1/6*3/4 + 1/2*5/6*1/4) + (1/2*1/6*3/4 + 1/2*5/6*1/4 + 1/2*1/6*1/4) + 1/2*1/6*1/4 =

= 11/16.

Answer: D.

zahiseif2 · Answer

hello, 
if the question is :     If    exactly        one of these three tasks is successful, then you win the game. What is the probability of winning the game?   instead of   If any of these task are successful, then you win the game. What is the probability of winning?

can we still use the same approach ? or we can just take the probability of 1 happening and other 2 no , and multiply the 3 to get 23/48 ?

Krunaal · Answer

If it is exactly one, then you calc. probability for three cases, in which each of the one different event is successful and two are unsuccessful, multiply and get probability of that case, and then add the probabilities of those 3 cases.

The other approach would be 1 - (Probability of exactly 2 wins + Probability of exactly 3 wins + Probability of 0 wins)

Kinshook · Answer

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.

The probability of NOT getting a heads on flipping a quarter = 1/2
The probability of NOT getting a six on rolling a single die = 5/6
The probability of NOT picking a spades card on picking a card from a full playing-card deck = 3/4

The probability of NOT getting success =  \frac{1}{2}*\frac{5}{6}*\frac{3}{4} = \frac{5}{16} 
The probability of getting success =  1 - \frac{5}{16} = \frac{11}{16}

IMO D

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

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