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13 Dec 2016, 07:18
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B
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Difficulty:
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92% (00:23) correct
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What is the probability of flipping a fair coin two times and the coin landing on heads on both flips?
A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 3/4
A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 3/4
10 Feb 2020, 08:44
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Bunuel
What is the probability of flipping a fair coin two times and the coin landing on heads on both flips?
A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 3/4
A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 3/4
Approach #1: Probability rules
P(both heads) = P(1st toss is heads AND 2nd toss is heads)
= P(1st toss is heads) x P(2nd toss is heads)
= 1/2 x 1/2
= 1/4
Answer: B
Approach #2: Counting
P(both heads) = (# of outcomes in which both tosses are heads)/(total # of possible outcomes)
Since there are so few outcomes, let's just list them:
- 1st toss is heads and 2nd toss is heads
- 1st toss is heads and 2nd toss is tails
- 1st toss is tails and 2nd toss is heads
- 1st toss is tails and 2nd toss is tails
total # of possible outcomes = 4
Among those 4 possible outcomes, only 1 outcome meets the condition that both tosses are heads
So, P(both heads) = 1/4
Answer: B
Cheers,
Brent
General Discussion
15 Dec 2016, 12:29
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Bunuel
What is the probability of flipping a fair coin two times and the coin landing on heads on both flips?
A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 3/4
A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 3/4
Probability of head on first flip = \(\frac{1}{2}\)
Probability of head on second flip = \(\frac{1}{2}\)
Thus, probability of flipping a fair coin two times and the coin landing on heads on both flips = \(\frac{1}{2}\)*\(\frac{1}{2}\) =>\(\frac{1}{4}\)
Hence, answer will be (B) \(\frac{1}{2}\)
