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Ram and Shyam planned to play a game of tossing coin. Each got chance
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07 Dec 2019, 18:03
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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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Re: Ram and Shyam planned to play a game of tossing coin. Each got chance
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07 Dec 2019, 19:00
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
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08 Dec 2019, 21:07
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
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07 Dec 2019, 23:51
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
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08 Dec 2019, 00:13
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
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08 Dec 2019, 01:20
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
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08 Dec 2019, 02:48
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^nr, 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
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08 Dec 2019, 05:05
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
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08 Dec 2019, 05:35
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
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08 Dec 2019, 12:54
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
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08 Dec 2019, 19:28
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
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09 Dec 2019, 00:52
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
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09 Dec 2019, 04:13
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
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09 Dec 2019, 20:34
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
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03 Jan 2020, 08:12
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 case1: HHHXX (1st 3 heads + last 2 don't care) P(HHHXX) = (1/2)^3*1^2=1/8 winning case2: THHHX P(THHHX) = (1/2)*(1/2)^3*1=1/16 winning case3: 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




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