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10 Jan 2021, 07:14
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B
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Difficulty:
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Question Stats:
49% (02:13) correct
51%
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wrong
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A box contains three coins, two of them are fair and the third is two-headed. A coin is selected at random and tossed. If heads appears, the coin is tossed again. If tails appears, then another coin is selected from the two remaining coins and tossed. What is the probability that heads appears twice?
A. 2/3
B. 1/2
C. 3/8
D. 1/4
E. 1/8
Are You Up For the Challenge: 700 Level Questions
A. 2/3
B. 1/2
C. 3/8
D. 1/4
E. 1/8
Are You Up For the Challenge: 700 Level Questions
Thegmatboy
GMAT 4: 740 Q47 V45
Posts: 31
10 Jan 2021, 07:29
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Unfair tossed twice= 1/3x 1 x 1
Fair tossed twice= 1/3x1/2x1/2 x2 for both instances
1/3+1/6=1/2
Answer is B
Rhys
Posted from my mobile device
Fair tossed twice= 1/3x1/2x1/2 x2 for both instances
1/3+1/6=1/2
Answer is B
Rhys
Posted from my mobile device
13 Dec 2021, 02:09
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Each coin being selected is an equally likely outcome. Therefore, the Probability of selecting any one coin is (1/3)
Coin A Fair: P(H) = 1/2
Coin B Fair: P(H) = 1/2
Coin C Loaded: P(H) = 2/2 = 1
Based on my understanding of the game:
You randomly pick a coin.
If Heads turns up on the first flip, you flip the same coin again. Whether you get Heads or Tails on this second flip, the game ends.
If Tails turns up on the first flip, then you flip ANOTHER coin one time. Whether you get Heads or Tails on this second flip, the game ends.
Therefore, the only way we can have Two Heads appear in the game is if we pick our first coin, Flip Heads, and the Flip Heads again. Any other result will not have 2 Heads appear.
Expected Value of the "weighted" probability: Each coin is equally likely to show up as our first picked coin.
(1/3)rd of the Time you would pick Coin A:
P(Heads on A) * P(Heads on A) = (1/2) (1/2) = 1/4
+
(1/3)rd of the Time you would pick Coin B:
P(Heads on B) * P(Heads on B) = (1/2) (1/2) = 1/4
+
(1/3)rd of the Time you would pick Coin C, the unfair coin. Since every Flip will turn up Heads, the Probability of getting Heads on 2 Flips = 1
(1/3) (1/4) + (1/3) (1/4) + (1/3)(1) =
(1/12) + (1/12) + (4/12) = (6/12) =
1/2
Coin A Fair: P(H) = 1/2
Coin B Fair: P(H) = 1/2
Coin C Loaded: P(H) = 2/2 = 1
Based on my understanding of the game:
You randomly pick a coin.
If Heads turns up on the first flip, you flip the same coin again. Whether you get Heads or Tails on this second flip, the game ends.
If Tails turns up on the first flip, then you flip ANOTHER coin one time. Whether you get Heads or Tails on this second flip, the game ends.
Therefore, the only way we can have Two Heads appear in the game is if we pick our first coin, Flip Heads, and the Flip Heads again. Any other result will not have 2 Heads appear.
Expected Value of the "weighted" probability: Each coin is equally likely to show up as our first picked coin.
(1/3)rd of the Time you would pick Coin A:
P(Heads on A) * P(Heads on A) = (1/2) (1/2) = 1/4
+
(1/3)rd of the Time you would pick Coin B:
P(Heads on B) * P(Heads on B) = (1/2) (1/2) = 1/4
+
(1/3)rd of the Time you would pick Coin C, the unfair coin. Since every Flip will turn up Heads, the Probability of getting Heads on 2 Flips = 1
(1/3) (1/4) + (1/3) (1/4) + (1/3)(1) =
(1/12) + (1/12) + (4/12) = (6/12) =
1/2
