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Hello punit2020

I have moved the question to the right forum and formatted it. Please provide us with the answer options so that the question can be presented in its original form. For the time being, I have added answer options, but they are not the official ones.

Thanks
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The options are not visible. The question appeared on the Offical mock no 3 of GMAT focus. Thank you for editing the question. I believe the options are correct.
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GMATCoachBen @ScottTargetTestPrep
Could you help with this question?­
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chetan2u

punit2020
What will be the answer for the above question if it was aksed that " What is the probability that Mardea will win this game?"

I apologize beforehand because i am not following the laid down guidelines for posting any question. I have my exam in a few days, running short on time.

Mardea has to score 3 or more.
Let there be 10 cards, out of which 3 are draw again. So these cards will have no role on winning or losing.

Mardea will have to finally end the game by picking any of 7 cards: FOUR cards of 2 points, TWO cards of 3 points and ONE card of 4 points.

If he picks any of 3s and 4, he will win so has to pick any of three cards out of seven cards => P = 3/7

Original question: There are THREE draw again cards. If any is picked, Mardea will have to pick another card. So probability is 3 in total 10 cards.
­

Hello. In this alternate question solution you said that the draw again cards will play no role. But what if Mardea picks up a draw again card and then he goes again?  And then maybe he picks up the draw again card again and so on. Wont that change the probability? I havent done the math this is just a thought that came in my mind. Maybe am complicating it too much?
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Okay, let me try,

Requirement: Mardea should draw atleast 2 times to win the game (i.e. scoring >= 3)

Now, thinking of the ways, she'd be in a position to draw cards atleast 2 times is when,

1. She draws a card, that says "draw again" & then she draws any other valid cards.
2. She draws a card, that says "score 2 points" & then tries her luck in subsequent tries.

Now translating this to math,

Probability(drawing atleast twice) = Possibility 1 + Possibility 2
= (3/10) + (7/10 * 4/7)
= 7/10.

punit2020


In a certain game, each player's turn consists of drawing a card from a randomized deck and immediately carrying out the instructions on the card. For example, if the card's instruction is to draw again, the player immediately draws another card and carries out that card's instruction. After the instruction is carried out, the card is removed from the game, and, except when the card instruction is to draw again, the player's turn ends. It is currently Mardea's turn, and she knows that each of the remaining cards instructs the player to do exactly one of the following: score 2 points, score 3 points, score 4 points, or draw again. Mardea will win the game this turn if and only if she scores 3 or more points on this turn. The diagram shows the probabilities associated with the possible outcomes for the next card that is drawn.

Select from the drop-down menus the options that complete the statement so that it is accurate based on the information provided.

The probability that Mardea will draw at least 2 cards this turn is in

Show SpoilerMy question
What will be the answer for the above question if it was aksed that " What is the probability that Mardea will win this game?"

I apologize beforehand because i am not following the laid down guidelines for posting any question. I have my exam in a few days, running short on time.

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The only scenario in which Mardea can draw two cards is when she gets a "Draw Again" on the first attempt. In a way, the question is asking the probability of "draw again" which is 3/10.
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punit2020
What will be the answer for the above question if it was aksed that " What is the probability that Mardea will win this game?"

I apologize beforehand because i am not following the laid down guidelines for posting any question. I have my exam in a few days, running short on time.


Mardea has to score 3 or more.
Let there be 10 cards, out of which 3 are draw again. So these cards will have no role on winning or losing.

Mardea will have to finally end the game by picking any of 7 cards: FOUR cards of 2 points, TWO cards of 3 points and ONE card of 4 points.

If he picks any of 3s and 4, he will win so has to pick any of three cards out of seven cards => P = 3/7

Original question: There are THREE draw again cards. If any is picked, Mardea will have to pick another card. So probability is 3 in total 10 cards.

Hi, I dont get it,

Why is it not 7/10*2/7 + 7/10*1/7 = 3/10?

why are we not considering all the cards and excluding just the draw again cards??
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chetan2u
punit2020
What will be the answer for the above question if it was aksed that " What is the probability that Mardea will win this game?"

I apologize beforehand because i am not following the laid down guidelines for posting any question. I have my exam in a few days, running short on time.


Mardea has to score 3 or more.
Let there be 10 cards, out of which 3 are draw again. So these cards will have no role on winning or losing.

Mardea will have to finally end the game by picking any of 7 cards: FOUR cards of 2 points, TWO cards of 3 points and ONE card of 4 points.

If he picks any of 3s and 4, he will win so has to pick any of three cards out of seven cards => P = 3/7

Original question: There are THREE draw again cards. If any is picked, Mardea will have to pick another card. So probability is 3 in total 10 cards.

Hi, I dont get it,

Why is it not 7/10*2/7 + 7/10*1/7 = 3/10?

why are we not considering all the cards and excluding just the draw again cards??



Hello !
We're looking for the probability that Mardea will draw at least 2 cards THIS TURN.

There are 4 different scenarios here :
1- Scores 2 points : Doesn't win but the player's turn ends -> 1 DRAW
2- Scores 3 points : Win the game -> 1 DRAW
3- Scores 4 points : Win the game -> 1 DRAW
4- Has to draw again -> AT LEAST 2 DRAW

So the probability to draw at least 2 cards is simply the "draw again" probability which is 3/10.
Dropdown 1: 3. Dropdown 2: 10.

That's how I understood this problem.
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I had a question. It says that she will only win if she gets 3 or more points. So wouldnt the prob of her winning and only drawing once be be 7/10*2/7 or 7/10*1/7 = 3/7. Then 1-3/7 would be prob of drawing 2 or more?
Fish181
In the spirit of DI questions, it made it really long and convoluted just to throw you off. It's simply asking given the probabilities, what are the odds that Mardea will draw 2 or more cards during her turn?

Well, she has to draw at least one card, so there is a 100% chance of that. We also know the only way to have more than one turn is if she gets the "draw again" card.

Knowing the rule of collectively exhaustive events https://www.statology.org/collectively-exhaustive/ we know that:

Probability of At least 2 = 1 - Probability(Only 1)

This is the mathematical way of saying "the odds of at least 2 turns is the total probability of anything possible (which is a 100% probability) minus the probability of only 1 turn."

The probability of only 1 turn would be if she scored points on the first draw, which the graph says is a 7/10 chance.

So the answer is 1 - 7/10 = 3/10 or 3 and 10.­
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Exactly, how I understood the problem, I don't know why people have over-complicated a simple problem, it's a 20 sec problem, and people are wasting 2 min 28 secs on it
alex384921







Hello !
We're looking for the probability that Mardea will draw at least 2 cards THIS TURN.

There are 4 different scenarios here :
1- Scores 2 points : Doesn't win but the player's turn ends -> 1 DRAW
2- Scores 3 points : Win the game -> 1 DRAW
3- Scores 4 points : Win the game -> 1 DRAW
4- Has to draw again -> AT LEAST 2 DRAW

So the probability to draw at least 2 cards is simply the "draw again" probability which is 3/10.
Dropdown 1: 3. Dropdown 2: 10.

That's how I understood this problem.
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The question asks 'The probability that Mardea will draw at least 2 cards this turn is' ______.
Atleast 2 means 2 or more. Now, she can also draw more than two here right? If she draws a 'draw again card' (P=3/10) and upon drawing again she draws a 'draw again' for another time (P=2/9). And this can happen once more because there is another 'draw again' card remaining in the deck (P=1/8).

Can someone clarify why its simply 3/10 thats the answer ?
punit2020


In a certain game, each player's turn consists of drawing a card from a randomized deck and immediately carrying out the instructions on the card. For example, if the card's instruction is to draw again, the player immediately draws another card and carries out that card's instruction. After the instruction is carried out, the card is removed from the game, and, except when the card instruction is to draw again, the player's turn ends. It is currently Mardea's turn, and she knows that each of the remaining cards instructs the player to do exactly one of the following: score 2 points, score 3 points, score 4 points, or draw again. Mardea will win the game this turn if and only if she scores 3 or more points on this turn. The diagram shows the probabilities associated with the possible outcomes for the next card that is drawn.

Select from the drop-down menus the options that complete the statement so that it is accurate based on the information provided.

The probability that Mardea will draw at least 2 cards this turn is in

Show SpoilerMy question
What will be the answer for the above question if it was aksed that " What is the probability that Mardea will win this game?"

I apologize beforehand because i am not following the laid down guidelines for posting any question. I have my exam in a few days, running short on time.

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This is a fairly easy question. Lets assume question asks us the probability of winning (i.e scoring 3 or more points). How do we solve then. Experts, please help.
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This is a fairly easy question. Lets assume question asks us the probability of winning (i.e scoring 3 or more points). How do we solve then. Experts, please help.
If the first card is draw again, he draws again. Rather he keeps drawing until he gets score card. So, ‘draw again’ cards do not have any role in winning.

knowing now that the result depends on score cards, M will win if she gets at least 3. So the answer becomes addition of probability of getting 3 and getting 4 => 2/7 + 1/7 = 3/7
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