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

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 .

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)

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)

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

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

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

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

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​​.

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

Posted from my mobile device

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.

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

= 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.