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Kevin flips a coin four times. what is the probability that
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05 Nov 2016, 06:10
05 Nov 2016, 06:10
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Question Stats:
91% (00:35) correct
8% (00:48) wrong based on 23 sessions
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Kevin flips a coin four times. what is the probability that he gets heads on at least one of the four flip?
a) \(\frac{1}{16}\)
b) \(\frac{1}{4}\)
c) \(\frac{3}{4}\)
d) \(\frac{13}{16}\)
e) \(\frac{15}{16.}\)
books says its option 'e' but i say its 'b'
what do you think?
justifiction:
[p(H)*p(T)*p(T)*p(T)] + [p(H)*p(H)*p(T)*p(T)] + [p(H)*p(H)*p(H)*p(T)] + [p(H)*p(H)*p(H)*p(H)]
=> [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] = 4/16 => 1/4.
thank you
siva
a) \(\frac{1}{16}\)
b) \(\frac{1}{4}\)
c) \(\frac{3}{4}\)
d) \(\frac{13}{16}\)
e) \(\frac{15}{16.}\)
books says its option 'e' but i say its 'b'
what do you think?
justifiction:
[p(H)*p(T)*p(T)*p(T)] + [p(H)*p(H)*p(T)*p(T)] + [p(H)*p(H)*p(H)*p(T)] + [p(H)*p(H)*p(H)*p(H)]
=> [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] = 4/16 => 1/4.
thank you
siva
Re: what is the answer 1/4 or 15/16
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05 Nov 2016, 08:06
05 Nov 2016, 08:06
2
Expert Reply
Hi,
The only case where we do not have a head is the case TTTT.
P(TTTT)=\(\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\)=\(\frac{1}{16}\).
So P(At least one head)= \(1 -\frac{1}{16}=\frac{15}{16}\).
Hence E is correct option.
The only case where we do not have a head is the case TTTT.
P(TTTT)=\(\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\)=\(\frac{1}{16}\).
So P(At least one head)= \(1 -\frac{1}{16}=\frac{15}{16}\).
Hence E is correct option.
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Kevin flips a coin four times. what is the probability that
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02 Oct 2022, 03:18
02 Oct 2022, 03:18
1
P(0H) + P(1H) + P(2H) + P(3H) + P(4H) = 1
=> P(1H) + P(2H) + P(3H) + P(4H) = 1 - P(0H)
P(At least 1 H) = 1 - P(0H)
1 - P(4T) = 1 - \(\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\) = 1 - \(\frac{1}{16}\) = \(\frac{15}{16}\)
So, Answer will be E
Hope it helps!
There will be 4 combination for 1 H and 3 Tails
HTTT, THTT, TTHT, TTTH
For 2H there will be 6 combinations
HHTT, HTHT, HTTH, THHT, THTH, TTHH
For 3 H there will be 4 combinations
HHHT, HHTH, HTHH, THHH
For 4 Heads there will be 1 combinations
HHHH
So, we need to add probability for all the cases if we doing it this way.
Watch the following video to learn How to Solve Probability with Coin Toss Problems
=> P(1H) + P(2H) + P(3H) + P(4H) = 1 - P(0H)
P(At least 1 H) = 1 - P(0H)
1 - P(4T) = 1 - \(\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\) = 1 - \(\frac{1}{16}\) = \(\frac{15}{16}\)
So, Answer will be E
Hope it helps!
ssiva wrote:
[p(H)*p(T)*p(T)*p(T)] + [p(H)*p(H)*p(T)*p(T)] + [p(H)*p(H)*p(H)*p(T)] + [p(H)*p(H)*p(H)*p(H)]
=> [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] = 4/16 => 1/4.
siva
=> [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] + [1/2 * 1/2 *1/2*1/2] = 4/16 => 1/4.
siva
There will be 4 combination for 1 H and 3 Tails
HTTT, THTT, TTHT, TTTH
For 2H there will be 6 combinations
HHTT, HTHT, HTTH, THHT, THTH, TTHH
For 3 H there will be 4 combinations
HHHT, HHTH, HTHH, THHH
For 4 Heads there will be 1 combinations
HHHH
So, we need to add probability for all the cases if we doing it this way.
Watch the following video to learn How to Solve Probability with Coin Toss Problems
