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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
For one toss of a certain coin, the probability that the outcome is he
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30 Mar 2022, 05:13
30 Mar 2022, 05:13
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Question Stats:
42% (01:41) correct
57% (01:14) wrong based on 7 sessions
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For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
A. \((0.6)^5\)
B. \(2(0.6)^4\)
C. \(3(0.6)^4\)
D. \(4(0.6)^4(0.4) + (0.6)^5\)
E. \(5(0.6)^4(0.4) + (0.6)^5\)
A. \((0.6)^5\)
B. \(2(0.6)^4\)
C. \(3(0.6)^4\)
D. \(4(0.6)^4(0.4) + (0.6)^5\)
E. \(5(0.6)^4(0.4) + (0.6)^5\)
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Re: For one toss of a certain coin, the probability that the outcome is he
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07 Apr 2022, 03:09
07 Apr 2022, 03:09
Expert Reply
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Explanation
Given: For one toss of a certain coin, the probability that the outcome is heads is 0.6.
Asked: If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
The probability that the outcome will be heads at least 4 times = The probability that the outcome will be heads 4 times + The probability that the outcome will be heads 5 times = 5C4*(.6)^4*(.4) + 5C5*(.6)^5 = 5(.6)^4(.4) + (.6)^5
IMO E
Given: For one toss of a certain coin, the probability that the outcome is heads is 0.6.
Asked: If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
The probability that the outcome will be heads at least 4 times = The probability that the outcome will be heads 4 times + The probability that the outcome will be heads 5 times = 5C4*(.6)^4*(.4) + 5C5*(.6)^5 = 5(.6)^4(.4) + (.6)^5
IMO E
Re: For one toss of a certain coin, the probability that the outcome is he
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13 Aug 2022, 06:16
13 Aug 2022, 06:16
1
Since there are only two possible outcomes, we can apply binomial probability distribution.
success = p = 0.6
failure = q = 0.4
P(at least 4 heads) = P(4 heads) + P(5 heads)
= 5C4\(p^4\)*\(q^1\) + 5C5\(p^5\)*\(q^0\)
= 5\((0.6)^4\)(0.4) + \((0.6)^5\)
success = p = 0.6
failure = q = 0.4
P(at least 4 heads) = P(4 heads) + P(5 heads)
= 5C4\(p^4\)*\(q^1\) + 5C5\(p^5\)*\(q^0\)
= 5\((0.6)^4\)(0.4) + \((0.6)^5\)
