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Ram and Shyam planned to play a game of tossing coin. Each got chance

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Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   07 Dec 2019, 19:03
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GMATBusters’ Quant Quiz Question -3


Ram and Shyam planned to play a game of tossing coin. Each got chance to toss the coin 5 times. The one who get 3 or more heads on consecutive tosses win. Ram started the game. What is the probability that Ram got 3 or more heads on consecutive tosses in first attempt?

A 2/16
B 1/4
C 7/24
D 5/16
E 15/32
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B
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   07 Dec 2019, 20:00
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Answer:B (1/4)

There are two outcomes for every flip (2)

As there are five tosess it is 2x2x2x2x2=32 (total outcomes)

there are 8 possibilities where we could get three or more heads consecutively

below are the combinations:
h h h h h
h h h h t
h h h t h
h t h h h
t h h h h
h h h t t
t h h h t
t t h h h

probability = favorable outcomes/total outcomes
=8/32=1/4
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   08 Dec 2019, 22:07
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The wording of the question is misleading. "What is the probability that Ram got 3 or more heads on consecutive tosses?" amounts to asking "What is the probability that Ram wins?" It doesn't ask "What is the probability that Ram got 3 or more heads on consecutive tosses in the first attempt?"
When someone wins, the game ends. If Ram gets 3 or more heads on consecutive tosses, he will win since he starts the game.

Let's first find the probability of getting 3 or more heads.
3 Heads - HHHTT, THHHT, TTHHH. We can get it in 3!/2! ways = 3 ways (because we have to arrange three elements T, T and HHH of which the two Ts are identical)
4 Heads - HHHHT, HHHTH, THHHH, HTHHH. We can get it in 4 ways because T has 4 distinct places in which it can be placed. It cannot be between two pairs of Hs each because 3 Heads need to be together.
5 Heads - HHHHH - We can get this in 1 way.

The total possible outcomes of 5 coin tosses are 2^5 = 32.

P(Three consecutive Heads) = (3 + 4 + 1)/32 = 1/4


Possibility of Ram winning = (1/4) + (3/4)*(3/4)*(1/4) + (3/4)^4*(1/4) + ...
Note how we get this:
(1/4) - Ram throw the die 5 times and gets 3 consecutive heads. Wins!
(3/4)*(3/4)*(1/4) - Ram throws the die 5 times but doesn't win. Shyam throws the die 5 times but doesn't win. Ram again throws and wins.
and so on...

This is an infinite GP.
Sum = (1/4) / (1 - 9/16) = 4/7
As per the question, this should be the answer.
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   08 Dec 2019, 00:51
Ans:B
Ram's Outcome : _ _ _ _ _

Total combination 2^5
Winning situations: a) Three consecutive Heads
(i) position(1,2,3)Head , (4) Tail and (5) Head or Tail = 2 arrangement
(ii)position(2,3,4)Head and (1,5) Tail = 1 arrangement
(iii)position(1)Head or Tail , (2) Tail and (3,4,5) head = 2 arrangement
b) Four consecutive heads
(i) Position(1,2,3,4) Head and (5) Tail = 1 arrangement
(ii)Position(1)Tail and(2,3,4,5) Head = 1 arrangement
c) Five consecutive Heads = 1 arrangement
Hence,
Winning Probablity= Probablity of 3 Heads+Probablity of 4 Heads+Probablity of 5 Heads
(2+1+2/2^5)+(2/2^5)+(1/2^5)
=1/4
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   08 Dec 2019, 01:13
all 5 heads = HHHHH ; 1
4 consecutive heads ; THHHH or HHHHT ; 2
3 consecutive heads ; TTHHH , THHHT, HHHTT , HTHHH, HHHTH ; 5
total possible cases as per condition; 1+2+5 ; 8
total possible cases; 2^5 ; 32
P = 8/32 ;1/4
IMO B

Ram and Shyam planned to play a game of tossing coin. Each got chance to toss the coin 5 times. The one who get 3 or more heads on consecutive tosses win. Ram started the game. What is the probability that Ram got 3 or more heads on consecutive tosses?

A 2/16
B 1/4
C 7/24
D 5/16
E 15/32
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   08 Dec 2019, 02:20
if we see, the denominator is 2^5 = 32

for numerator only following 8 cases are possible
HHHTT
THHHT
TTHHH

HHHHT
HHHTH
HTHHH
THHHH

HHHHH

So answer is 8/32=1/4
B
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   08 Dec 2019, 03:48
Ram and Shyam planned to play a game of tossing coin. Each got chance to toss the coin 5 times. The one who get 3 or more heads on consecutive tosses win. Ram started the game. What is the probability that Ram got 3 or more heads on consecutive tosses?

A 2/16
B 1/4
C 7/24
D 5/16
E 15/32

Multiple trials of a single event: If multiple independent trials of a single event are performed, then the probability of r successes out of a total of n trials can be determined = nCr * p^r * q^n-r,
where,
n = number of times the event is performed = 5
r = number of successes = 3 consecutive
p = probability of success in one trial = 1/2
q = probability of failure in one trial = 1/2

5C3 * 1/2^3 * 1/2^2
5!/3!*2! *1/8 * 1/4
5/16.

Imo. D
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   08 Dec 2019, 06:05
The answer is B: 1/4.

EXPLANATION:
Let's find the total number of cases, which is 2^5 = 32 (as 5 coins are tossed, each having the chance of having either heads or tails, or two outcomes).

Now, we find the probable number of cases.

Given constraint: 3 or more consecutive heads.

Let's form the cases.

Case 1: 3H, 2T
Total arrangement possible: 3
3H 2T, T 3H T, 2T 3H

Case 2: 4H, 1T
Total arrangement possible: 2
4H 1T, 1T 4H.

Case 3: 5H
Total arrangement possible: 1

Case 4: 3H T H
Total arrangement possible: 2
3H T H, H T 3H

So total number of cases: 3 + 2 + 1 + 2 = 8

Hence, probability = total number of valid cases/total cases = 8/32 = 1/4, which is the answer.

Hence, answer is B: 1/4.
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   08 Dec 2019, 06:35
Method - I
Probability of getting a head, P (H) = 1/2
Probability of getting a tail, P (T) = 1/2
Probability of getting a head or a tail, P (H/T) = 1

For Ram to get 3 Heads, following 3 cases are possible

1. First three consecutive heads, remaining two head/tail.
H H H (H/T) (H/T)
P = (1/2)^3*1*1 = 1/8

2. First tail, then three consecutive heads, followed by head/tail.
T H H H (H/T)
P = (1/2)*(1/2)^3*1 = 1/16

3. First head/tail, then tail, followed by three consecutive heads.
(H/T) T H H H
P = 1*(1/2)*(1/2)^3 = 1/16

Therefore, total probability = 1/8 + 1/16 + 1/16 = 1/4

Method – II

At least 3 consecutive heads means 3, 4 or 5 consecutive heads

For 3 consecutive heads
5 cases are possible:
H H H T T
H H H T H
T H H H T
T T H H H
H T H H H

P = 5 * (1/2)^5 = 5/32

For 4 consecutive heads
2 cases are possible:
H H H H T
T H H H H

P = 2 * (1/2)^5 = 2/32

For 5 consecutive heads
1 case is possible:
H H H H H

P = (1/2)^5 = 1/32
Total Probability = 5/32 + 2/32 + 1/32 = 8/32 = 1/4
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   08 Dec 2019, 13:54
By simple counting there are 5 ways to get 3 consecutive heads
HHHTT
TTHHH
THHHT
HTHHH
HHHTH We will multiply 1/32 by 5 to get this probability
There are 2 ways to get 4 consecutive heads:
HHHHT
THHHH. We will multiply 1/32 by 2 to get this probability

Finally there is is 1 way to get 5 heads : HHHHH. we will multiply 1/32 by 1 to get this probability. Finally, we add all to get 1/4.


Answer is B
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   08 Dec 2019, 20:28
Hi! Can you explain the answer?
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   09 Dec 2019, 01:52
The are 4 options:
1) all 5 tosses heads - 1/32 probability
2) 4 tosses heads in a row - (1/2^4) * (1/2) * 2 = 2/32
The idea is that we can imagine that 4 consecutive tosses is one entity, so there are exactly 2 possibilities to have 4 heads in a row. HHHHT and THHHH.
3) 3 tosses in a row - (1/2^3) * (1/2^2) * 3 = 3/32
3 is because we can choose 1 out of 3 in 3 ways HHHTT, THHHT, TTHHH
4) We must take into account the situation in which we have 4 heads, but they are not consecutive: HTHHH, HHHTH
2/32

So we have 1/32 + 2/32 + 3/32 + 2/32 = 8/32= 1/4

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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   09 Dec 2019, 05:13
At least 3 consecutive heads implies 3, 4 or 5 consecutive heads

For 3 consecutive heads
5 cases:
HHHTT
HHHTH
THHHT
TTHHH
HTHHH

P = 5*(1/2)^5 = 5/32

For 4 consecutive heads
2 cases:
HHHHT
THHHH

P = 2*(1/2)^5 = 2/32

For 5 consecutive heads
1 case:
HHHHH

P = 1*(1/2)^5 = 1/32

Total P = 5/32 + 2/32 + 1/32 = 8/32 = 1/4

B is the answer.
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   09 Dec 2019, 21:34
Expert Reply
Hi Karishma

You are right, adding " in first attempt" makes the question more precise and unambiguous.

Thank you

VeritasKarishma wrote:
The wording of the question is misleading. "What is the probability that Ram got 3 or more heads on consecutive tosses?" amounts to asking "What is the probability that Ram wins?" It doesn't ask "What is the probability that Ram got 3 or more heads on consecutive tosses in the first attempt?"
When someone wins, the game ends. If Ram gets 3 or more heads on consecutive tosses, he will win since he starts the game.

Let's first find the probability of getting 3 or more heads.
3 Heads - HHHTT, THHHT, TTHHH. We can get it in 3!/2! ways = 3 ways (because we have to arrange three elements T, T and HHH of which the two Ts are identical)
4 Heads - HHHHT, HHHTH, THHHH, HTHHH. We can get it in 4 ways because T has 4 distinct places in which it can be placed. It cannot be between two pairs of Hs each because 3 Heads need to be together.
5 Heads - HHHHH - We can get this in 1 way.

The total possible outcomes of 5 coin tosses are 2^5 = 32.

P(Three consecutive Heads) = (3 + 4 + 1)/32 = 1/4


Possibility of Ram winning = (1/4) + (3/4)*(3/4)*(1/4) + (3/4)^4*(1/4) + ...
Note how we get this:
(1/4) - Ram throw the die 5 times and gets 3 consecutive heads. Wins!
(3/4)*(3/4)*(1/4) - Ram throws the die 5 times but doesn't win. Shyam throws the die 5 times but doesn't win. Ram again throws and wins.
and so on...

This is an infinite GP.
Sum = (1/4) / (1 - 9/16) = 4/7
As per the question, this should be the answer.
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   03 Jan 2020, 09:12
Ram and Shyam planned to play a game of tossing coin. Each got chance to toss the coin 5 times. The one who get 3 or more heads on consecutive tosses win. Ram started the game. What is the probability that Ram got 3 or more heads on consecutive tosses in first attempt?

A 2/16
B 1/4 --> correct
C 7/24
D 5/16
E 15/32

Solution:
winning case-1: HHHXX (1st 3 heads + last 2 don't care)
P(HHHXX) = (1/2)^3*1^2=1/8
winning case-2: THHHX
P(THHHX) = (1/2)*(1/2)^3*1=1/16
winning case-3: XTHHH
P(XTHHH) = (1)*(1/2)*(1/2)^3=1/16

Total probability = 1/8+1/16+1/16=(2+1+1)/16=4/16=1/4
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Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   29 Mar 2022, 04:56
VeritasKarishma wrote:

Possibility of Ram winning = (1/4) + (3/4)*(3/4)*(1/4) + (3/4)^4*(1/4) + ...
Note how we get this:
(1/4) - Ram throw the die 5 times and gets 3 consecutive heads. Wins!
(3/4)*(3/4)*(1/4) - Ram throws the die 5 times but doesn't win. Shyam throws the die 5 times but doesn't win. Ram again throws and wins.
and so on...

This is an infinite GP.
Sum = (1/4) / (1 - 9/16) = 4/7
As per the question, this should be the answer.



Didn't understand from this part. Could you please explain.

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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink] New post   17 Jul 2023, 20:24
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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance [#permalink]
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