GATE2018 ME-2: GA-10

Question

An unbiased coin is tossed six times in a row and four different such trials are conducted. One trial implies six tosses of the coin. If H stands for head ans T stands for tail, the following are the observations from the four trials.

1. HTHTHT
2. TTHHHT
3. HTTHHT
4. HHHT_ _

Which statement describing the last two coin tosses of the fourth trial has the highest probability of being correct?

• A. Two T will occur.
• B. One H and one T will occur.
• C. Two H will occur.
• D. One H will be followed by one T.

Comments

It’s clearly given “AN UNBIASED COIN”, so no need to check previous trials.If it was biased , we could consider it.

Answer 1

The outcome of a coin toss is independent of the previous outcomes - so no need to look into history.

Possible Outcomes:

1. $H H$
2. $H T$
3. $T H$
4. $T T$

• A. $P(2T) = P(T T) = 1/4 = 0.25$
• B. $P(1H \text{ AND } 1T) = P(H T) + P(T H) = 1/4 + 1/4 =0.25 + 0.25 = 0.5$
• C. $P(2H) = P(H H) = 1/4 = 0.25$
• D. $P(H T) = 1/4 = 0.25$

$0.5 > 0.25,$ So B is Correct 😊

Comments

Wow Balaji ! :)

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You know its very simple intuition also, that getting head or tail in next event is independent of what has happened earlier to coin, so simply at each place either Head or Tail will occur.