swerve 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

The OA is E

Source: Magoosh

Say

the probability of getting a head = p = 1/2; thus, the probability of getting a tail = q = 1/2;
the probability of getting a six = x = 1/6; thus, the probability of getting any among (1, 2, 3, 4 and 5) = y = 5/6;
the probability of getting a card of spade = r = 13/52 = 1/4; the probability of getting a card other than spade = s = 39/52 = 3/4;

Given that if exactly one of these three tasks is successful, then you win the game.

The probability of winning = pys + qxs + qyr = 1/2*5/6*3/4 + 1/2*1/6*3/4 + 1/2*5/6*1/4 = 23/48

The correct answer: E

Hope this helps!

-Jay
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swerve 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

The OA is E

Source: Magoosh

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

Hi All,

We're told that 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. We're told that if you perform EXACTLY ONE of these three tasks is successful, then you win the game (meaning that any other outcomes - re: 0, 2 or 3 successes would NOT be a "win"). We're asked for the probability of winning the game.

Since there are 2 outcomes with a coin, 1/2 of the outcomes are successes and 1/2 are not
Since there are 6 outcomes with rolling a die, 1/6 of the outcomes are successes and 5/6 are not
Since there are 52 outcomes with choosing a card (and 13 cards are spades), 1/4 of the outcomes are successes and 3/4 are not

There are 3 ways to 'win' the game:
1) Win the coin, Lose the die, Lose the card = (1/2)(5/6)(3/4) = 15/48
2) Lose the coin, Win the die, Lose the card = (1/2)(1/6)(3/4) = 3/48
3) Lose the coin, Lose the die, Win the card = (1/2)(5/6)(1/4) = 5/48

Total probability of winning the game = 15/48 + 3/48 + 5/48 = 23/48

Final Answer: [spoiler=]E[/spoiler]

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