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03 Sep 2022, 04:46
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
64% (01:18) correct
36%
(01:25)
wrong
based on 120
sessions
History
Date
Time
Result
Not Attempted Yet
When a coin is tossed for 6 times, what is the probability that after the first tossing, every outcome will be different from the previous one?
(A) 1/16
(B) 1/24
(C) 1/32
(D) 1/48
(E) 1/64
(A) 1/16
(B) 1/24
(C) 1/32
(D) 1/48
(E) 1/64
decentprashant
04 Sep 2022, 05:10
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P (Event) = Desired Outcome / Total Outcomes = 2/2^6 = 2/64 = 1/32
Desired outcome = P(HTHTHT) & P (THTHTH) = 2
Total Outcomes = 2^6 = 64
Desired outcome = P(HTHTHT) & P (THTHTH) = 2
Total Outcomes = 2^6 = 64
16 Sep 2022, 10:08
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Bunuel
Each throw can have total two outcomes , H or T
For the first throw desired outcomes are two as we can have any of H or T = so probability is \(\frac{2}{2}=1\)
For the second throw we can have only one option as it needs to be different from the first one , hence p = \(\frac{1}{2}\)
For the third throw we can have only one option as it needs to be different from the previous one , hence p = \(\frac{1}{2}\)
Similarly for fourth throw p will be = \(\frac{1}{2}\)
Similarly for fifth throw p will be = \(\frac{1}{2}\)
Similarly for the sixth throw p will be = \(\frac{1}{2}\)
Hence finally we get p = 1*\(\frac{1}{2}*\)*\(\frac{1}{2}\)*\(\frac{1}{2}\)*\(\frac{1}{2}\)*\(\frac{1}{2}\) =\(\frac{1}{32}\)
Ans C
Hope it's clear.
