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Re: minimum number of heads in the first 10 tosses [#permalink]
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akikap2609 wrote:
Please give the solution. I can't understand why it isn't C.


My bad. I mismatched the OA. It is C

OE

Quote:
Since there are 8 total heads and the objective is to have a minimal quantity of them in the first ten tosses, the last four tosses would have to all be heads. This would leave 8 − 4 = 4 tosses left for the first ten outcomes.
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Re: minimum number of heads in the first 10 tosses [#permalink]
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Carcass wrote:
A coin is tossed 14 times.
Heads occurs 8 times and tails 6 times.
The second and thirteenth tosses are heads.


Quantity A
Quantity B
minimum number of heads in the first 10 tosses
4



A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

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We have 8H and 6T
Also, the pattern is - _ H _ _ _ _ _ _ _ _ _ _ H _

In order to find the minimum number of Heads in the first 10 tosses, we have to maximize the Tails (max. 6) in the first 10 tosses
i.e. {T H T T T T T H H H} H H H H

Col. A: 4
Col B: 4

Col. A = Col. B
Hence, option C
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Re: minimum number of heads in the first 10 tosses [#permalink]
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There are total 8 H's and 6 T's
Restriction is 2nd and 13th spots are H.
Sequence will look sth like:

__ H __ __ __ __ __ __ __ __ __ __ H __
To minimise H's in the first 10 spots, maximise the T's.
Filling 6 out of first 10 spots with T's. Now we are out of T's, fill remaining 3 spots (one is already fixed as H) with H.

Total minimum 4 heads.
Ans C
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Re: minimum number of heads in the first 10 tosses [#permalink]
Given that A coin is tossed 14 times. Heads occurs 8 times and tails 6 times. The second and thirteenth tosses are heads.

Quantity A: Minimum number of heads in the first 10 tosses

As we need to find the minimum number of heads in the first toss and we already know that the second toss is 2, so we will try to get as many number of tails as possible in the first 10 toss.

We know that there are 6 Tails.
So, all 6 will occur in first 10 toss and we already know that 2nd toss was head
=> Remaining number of heads = 10 - 6(for 6 tails) - 1(for 2nd toss which resulted in heads) = 3

=> Minimum number of heads in first 10 = 3 + 1 = 4

Clearly, Quantity A = Quantity B = 4

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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Re: minimum number of heads in the first 10 tosses [#permalink]
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