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 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.

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

iamdp · Answer

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.  A  BC  = 1/2*5/6*3/4 = 15/48
OR
2.  A  B  C  = 1/2*1/6*3/4 = 3/48
OR
3.  AB  C  = 1/2*5/6*1/4 = 5/48

Now as we now OR = +

P of winning = 15/48 + 3/48 + 5/48 = 23/48

23a2012 · Answer

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

Bunuel · Answer

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)

animesh_an · Answer

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

IanStewart · Answer

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

RiyazAhamed · Answer

Is it possible to do this question by 1 -  ?

It anyways signifies, any one task turning to be favorable. I tried doing this way, but ended up getting 11/16 and chose the wrong answer D. What am I missing?

bumpbot · Answer

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