Daniel is playing a game with a dice and a coin to win up to two prize

Question

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Daniel is playing a game with a dice and a coin to win up to two prizes.

The dice is numbered 1 through 6, and the coin has numbers 1 and 2. Daniel rolls the dice once and flips the coin once.

• If the product of the two numbers is even, he wins a smartwatch.

• If the dice and coin show the same number, he wins a laptop.

Select for  None  the probability that Daniel will not win any prize and for  Both prizes , the probability that Daniel will win both prizes.

Bunuel · Accepted Answer

Official Solution: For Daniel not to win any prize, he must: 1. Roll an odd number and get an odd number on the coin flip, as this is the only way to avoid an even product, and 2. Avoid the (1, 1) case (because in this case, the dice and coin show the same number, and he would win a laptop). Therefore, he must roll 3 or 5 on the die and get 1 on the coin flip. The probability of this is (2/6) * (1/2) = 1/6. For Daniel to win both prizes, he must get an even product and matching numbers. This is possible if he rolls 2 and flips 2. The probability of this is (1/6) * (1/2) = 1/12. Correct answer: None "1/6" Both Prizes "1/12" GMAT-Club-Forum-z4jgx85a.png

Schinchan_bh · Answer

There are 12 possible instances (1,1) (1,2)
 (2,1) (2,2)
 (3,1) (3,2)
 (4,1) (4,2)
 (5,1) (5,2)
 (6,1) (6,2)
to win none the composition is (3,1) and (5,1) only so 1/6
to win both (2,2) 1/12

Aryaa03 · Answer

Sol: 
IMO
P(None) = 2/12
P (Both) = 1/12

a) Sample Space for Coin = { 1, 2 } 
b) Sample Space for Dice = { 1, 2, 3, 4, 5, 6 }

Total no. of events with single roll of dice and single flip of coin = 6* 2

--Events resulting in even no. (Coin, Dice) => (2,1) , (2,2), (2,3), (2,4), (2,5) , (2,6) , (1,2) , (1,4) , (1,6) ==>No. of favourable events = 9
P(Even Prod) = 9/12

--Events with same no. (Coin, Dice) => (1,1) , (2,2) ==> No. of favourable events = 2
P(Same) = 2/12

--Both both above events, (2,2) is the common occurence
Therefore, P(Even and Same) = 1/12

For 1st Prompt: 
P(Neither) = 1 - P(Either) = 1 - { 9/12 + 2/12 - 1/12 } = 1 - 10/12 = 2/12

For 2nd Prompt: 
P(Both) = P(Even and Same) = 1/12

Ilanchezhiyan · Answer

I solved it like this;
 For Both: 
He has to get a 2 in the coin and 2 in the Dice; 
In probability terms, this looks like P(Getting a two in the dice) * P(Getting a two in the coin) = (1/6)*(1/2) = 1/12
 For None: 
He has to get Odd values except 1 in the Dice and one in the coin(two would make the product even);
P(Odd values in the Dice except one) * P(Getting one in the Coin) = (2/6)*(1/2)= 1/6

Aalto700 · Answer

1. None mean odd * odd and not combine 1-1

dice: 3, 5 - prob 2/6
coin: 1 - prob 1/2
 2/6 * 1/2 = 1/6

2. Both: 
1/6 *1/2 =1/12

Daniel is playing a game with a dice and a coin to win up to two prize

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Daniel is playing a game with a dice and a coin to win up to two prize

Prep Toolkit

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