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

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

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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

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BTGmoderatorDC 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 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
Source: Magoosh
$$\left. \matrix{
{\rm{i}}\,\,{\rm{:}}\,\,\,{\rm{get}}\,\,{\rm{H}}\,\,{\rm{in}}\,\,{\rm{quarter}} \hfill \cr
{\rm{ii}}:\,\,{\rm{get}}\,\,6\,\,{\rm{in}}\,{\rm{die}} \hfill \cr
{\rm{iii}}\,\,{\rm{:}}\,\,{\rm{get}}\,\,{\rm{spades}}\,{\rm{in}}\,{\rm{52}}\,\,{\rm{cards}}\,{\rm{regular}}\,\,{\rm{deck}} \hfill \cr} \right\}\,\,\,{\rm{independent}}\,\,{\rm{events}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{their}}\,\,{\rm{negations}}\,\,{\rm{are}}\,\,{\rm{independent}}\,\,{\rm{events}}\,\,\,\,\,\left( * \right)$$
$$? = P\left( {{\rm{i}}\,\,{\rm{OR}}\,\,{\rm{ii}}\,\,{\rm{OR}}\,\,{\rm{iii}}} \right) = 1 - P\left( {\left( {{\rm{not}}\,\,{\rm{i}}} \right)\,\,AND\,\,\left( {{\rm{not}}\,\,{\rm{ii}}} \right)\,\,AND\,\,\left( {{\rm{not}}\,\,{\rm{iii}}} \right)} \right)$$
$$P\left( {\left( {{\rm{not}}\,\,{\rm{i}}} \right)\,\,AND\,\,\left( {{\rm{not}}\,\,{\rm{ii}}} \right)\,\,AND\,\,\left( {{\rm{not}}\,\,{\rm{iii}}} \right)} \right)\,\,\mathop = \limits^{\left( * \right)} \,\,\,{1 \over 2} \cdot {5 \over 6} \cdot {{52 - 13} \over {52}} = {5 \over {16}}$$
$$? = 1 - {5 \over {16}} = {{11} \over {16}}$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or https://GMATH.com.br (Portuguese version)
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BTGmoderatorDC 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 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
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

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Scott Woodbury-Stewart Founder and CEO

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