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Sub 505 Level|   Probability|      
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If a certain coin is flipped, the probability that the coin will land 14 Feb 2020, 11:08
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marcodonzelli
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32
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The probability that the first 3 flips are heads and last 2 flips are tails is:

P(HHHTT) = (1/2)^5 = 1/32.

Answer: E
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If a certain coin is flipped, the probability that the coin will land 13 Apr 2022, 04:27
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marcodonzelli
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32

We want the probability that coin will land heads up on the first 3 flips and not on the last 2 flips. So there is ONLY one favorable outcome, namely heads up on the first 3 flips and tails up on the last 2 flips: HHHTT.

# of total out comes is 2^5=32.

P=favorable/total=1/32.

Answer: E.

IF THE QUESTION WERE:
A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)

Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\).

As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question.

Hope it's clear.
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Hi brunel, great info always. Thought the repitition is taken care of by 3!2! already. Therefore not sure why are we multiplying (1/2)^3∗(1/2)^2 again? Am I missing something? Could you explain? Thanks

P=5!/3!2!∗(1/2)^3∗(1/2)^2
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If a certain coin is flipped, the probability that the coin will land 13 Apr 2022, 18:08
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Bunuel
marcodonzelli
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32

We want the probability that coin will land heads up on the first 3 flips and not on the last 2 flips. So there is ONLY one favorable outcome, namely heads up on the first 3 flips and tails up on the last 2 flips: HHHTT.

# of total out comes is 2^5=32.

P=favorable/total=1/32.

Answer: E.

IF THE QUESTION WERE:
A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)

Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\).

As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question.

Hope it's clear.

Hi brunel, great info always. Thought the repitition is taken care of by 3!2! already. Therefore not sure why are we multiplying (1/2)^3∗(1/2)^2 again? Am I missing something? Could you explain? Thanks

P=5!/3!2!∗(1/2)^3∗(1/2)^2
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The (1/2)^3*(1/2)^2 is the probability of 3 heads and 2 tails occurring in one specified pattern or way, in 5 tosses.

5!/3!2! is the number of ways to create 3 heads and 2 tails in 5 tosses.

5! is the number of arrangements of 5 different things.

Since 3 heads are being treated above as if they were 3 different things with 3! arrangements, where in fact they are just one arrangement, the 5! needs to be divided by 3!. The same logic applies to the 2 tails.






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If a certain coin is flipped, the probability that the coin will land 15 Apr 2022, 13:10
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Brilliant and thanks Regor60. Understand now.
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If a certain coin is flipped, the probability that the coin will land Updated on: 29 Jul 2025, 07:37
Originally posted by BrushMyQuant on 05 Oct 2022, 07:49.
Last edited by BrushMyQuant on 29 Jul 2025, 07:37, edited 1 time in total.
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Given that If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, and we need to find what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

Coin is tossed 5 times => Total number of cases = \(2^5\) = 32

We need to get head in the first 3 flips and not in the last 2 flips => HHHTT

So, there is only one case out of 32 when this will happen.

=> P(HHHTT) = \(\frac{1}{32}\)

So, Answer will be E
Hope it helps!

Playlist on Solved Problems on Probability here

Watch the following video to MASTER Probability with Coin Toss Problems

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If a certain coin is flipped, the probability that the coin will land 27 May 2025, 19:44
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A conceptual doubt, in probability when do we use combinations and when do we use permutation.

Sorry for the dumb question, I am a little lost in P&C and prob.

Bunuel
marcodonzelli
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32

We want the probability that coin will land heads up on the first 3 flips and not on the last 2 flips. So there is ONLY one favorable outcome, namely heads up on the first 3 flips and tails up on the last 2 flips: HHHTT.

# of total out comes is 2^5=32.

P=favorable/total=1/32.

Answer: E.

IF THE QUESTION WERE:
A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)

Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\).

As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question.

Hope it's clear.
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If a certain coin is flipped, the probability that the coin will land Updated on: 21 Aug 2025, 06:55
Originally posted by Bunuel on 27 May 2025, 23:35.
Last edited by Bunuel on 21 Aug 2025, 06:55, edited 1 time in total.
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SanchitSanjay
A conceptual doubt, in probability when do we use combinations and when do we use permutation.

Sorry for the dumb question, I am a little lost in P&C and prob.


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