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16 Apr 2020, 04:30
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An unbiased coin (a coin which has an equal probability of heads and tails) is tossed 4 times, what is the probability that it shows three same results out of the 4 times?
A. 1/6
B. 1/8
C. 1/4
D. 1/2
E. 3/4
A. 1/6
B. 1/8
C. 1/4
D. 1/2
E. 3/4
Searched the forum for a similar ques before posting
16 Apr 2020, 04:37
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Kritisood
An unbiased coin (a coin which has an equal probability of heads and tails) is tossed 4 times, what is the probability that it shows three same results out of the 4 times?
A. 1/6
B. 1/8
C. 1/4
D. 1/2
E. 3/4
A. 1/6
B. 1/8
C. 1/4
D. 1/2
E. 3/4
Searched the forum for a similar ques before posting
So, we need any combination of TTTH (TTTH, TTHT, THTT, HTTT) and any combination of HHHT (HHHT, HHTH, HTHH, THHH).
P = 4C3*1/2^4 + 4C3*1/2^4 = 1/4 + 1/4 = 1/2.
Answer: D.
General Discussion
04 Sep 2020, 08:26
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\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)
When a coin is flipped n times, the total number of outcomes = \(2^n\)
When it is flipped 4 times, the total outcomes = \(2^4\) = 16
Favorable outcomes is when 3 out of the 4 flips is the same i.e we get 3 Heads and 1 Tail or 3 Tails and 1 Head
(a) 3 Heads and 1 Tail = 4C3 = 4
(b) 3 Tails and 1 Head = 4C3 = 4
Total Favorable outcomes = 4 + 4 = 8
Therefore the required probability = \(\frac{8}{16} = \frac{1}{2}\)
Option D
Arun Kumar
When a coin is flipped n times, the total number of outcomes = \(2^n\)
When it is flipped 4 times, the total outcomes = \(2^4\) = 16
Favorable outcomes is when 3 out of the 4 flips is the same i.e we get 3 Heads and 1 Tail or 3 Tails and 1 Head
(a) 3 Heads and 1 Tail = 4C3 = 4
(b) 3 Tails and 1 Head = 4C3 = 4
Total Favorable outcomes = 4 + 4 = 8
Therefore the required probability = \(\frac{8}{16} = \frac{1}{2}\)
Option D
Arun Kumar
