In a certain game, you perform three tasks. You flip a quarter, and success would be heads. You roll a single die, and

[BTGmoderatorDC]

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


OA D

Source: Magoosh

[swerve]

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


OA D

Source: Magoosh

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

So, Probability NO heads\(=\dfrac{1}{2}\)

Probability NO number \(6 =\dfrac{5}{6}\)

Probability NO picking a spades card \(=\dfrac{3}{4}\)

So, Probability NO winning \(= \dfrac{1}{2}\cdot \dfrac{5}{6}\cdot \dfrac{3}{4} = \dfrac{15}{48}=\dfrac{5}{16}\)

Probability NO winning \(= 1 - \dfrac{5}{16} = \dfrac{11}{16}\Longrightarrow\)D

[Scott@TargetTestPrep]

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


OA D

Solution:

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