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A biased coin's probability of landing on head is 2/5.
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Updated on: 08 Mar 2021, 00:41
Updated on: 08 Mar 2021, 00:41
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A biased coin's probability of landing on head is 2/5.
What is the probability of getting at least 1 tail, when the coin is flipped 4 times?
A: \(\frac{16}{625}\)
B: \(\frac{81}{625}\)
C: 1
D: \(\frac{609}{625}\)
E: \(\frac{544}{625}\)
What is the probability of getting at least 1 tail, when the coin is flipped 4 times?
A: \(\frac{16}{625}\)
B: \(\frac{81}{625}\)
C: 1
D: \(\frac{609}{625}\)
E: \(\frac{544}{625}\)
Originally posted by kvarunkumar1975 on 07 Mar 2021, 19:44.
Last edited by Carcass on 08 Mar 2021, 00:41, edited 1 time in total.
Last edited by Carcass on 08 Mar 2021, 00:41, edited 1 time in total.
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Re: A biased coin's probability of landing on head is 2/5.
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07 Mar 2021, 19:48
07 Mar 2021, 19:48
2
4Heads = 2/5*2/5*2/5*2/5
at least 1 tail = 1 - 4H = 609/625
at least 1 tail = 1 - 4H = 609/625
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Re: A biased coin's probability of landing on head is 2/5.
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07 Mar 2021, 23:02
07 Mar 2021, 23:02
kvarunkumar1975 wrote:
A biased coin's probability of landing on head is 2/5.
What is the probability of getting at least 1 tail, when the coin is flipped 4 times?
A: \(\frac{16}{625}\)
B: \(\frac{81}{625}\)
C: 1
D: \(\frac{609}{625}\)
E: \(\frac{544}{625}\)
What is the probability of getting at least 1 tail, when the coin is flipped 4 times?
A: \(\frac{16}{625}\)
B: \(\frac{81}{625}\)
C: 1
D: \(\frac{609}{625}\)
E: \(\frac{544}{625}\)
Probability of getting a Head = \(\frac{2}{5}\)
Probability of getting a Tail = \(1 - (\frac{2}{5}) = \frac{3}{5}\)
Now, Probability of getting at least 1 tail, when the coin is flipped 4 times = \(1 - Probability of No Tail\)
i.e. \(1 - (\frac{2}{5})^4\)
\(1 - \frac{16}{625} = \frac{609}{625}\)
Hence, option D
