When tossed, a certain coin has equal probability of landing

Winning scenario is if we'll have either three tails (TTT) or three heads (HHH):

\(\frac{1}{2}*\frac{1}{2}*\frac{1}{2}+\frac{1}{2}*\frac{1}{2}*\frac{1}{2}=\frac{1}{4}\).

Answer: B.
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There are 8 possibilities for outcomes when flipping a fair coin 3 times.

2*2*2=2^3=8

of these 8 possibilities there are 2 with the coin landing on the same side each time. HHH and TTT.

so the probability is 2/8 or 1/4.

HHH -- works
HHT
HTH
HTT
THH
THT
TTH
TTT -- works

the standard .5*.5*.5=1/8 but that's just the probability of getting one SPECIFIC side 3 times in a row (ie probability of 3 heads). Since they don't care if it's 3 heads or 3 tails you have to multiply 1/8 by 2 (so now it's the probability of 3 heads OR 3 tails)

1st throw: Probability of getting either side = 1
2nd throw: Probability of getting the first side = 1/2
3rd throw: Probability of getting the first side = 1/2

(1/2)*(1/2)=1/4

We need to determine the probability that a coin lands on the same side for all 3 tosses of the coin. This means that the 3-toss outcome will be either H-H-H or T-T-T.

P(T-T-T) = 1/2 x 1/2 x 1/2 = 1/8

P(H-H-H) = 1/2 x 1/2 x 1/2 = 1/8

Thus, the probability that the coin lands on the same side each time is 1/8 + 1/8 = 2/8 = 1/4.

Answer: B
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P(3 matching tosses) = P(ANY result on 1st toss AND 2nd toss matches 1st toss AND 3rd toss matches 2nd toss)
= P(ANY result on 1st toss) x P(2nd toss matches 1st toss ) x P(3rd toss matches 2nd toss)
= 1 x 1/2 x 1/2
= 1/4
= B

Cheers,
Brent

@Bunel, GMAT prep question. Please tag it.

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Done. Thank you!
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We need to find On three consecutive flips of a coin, what is the probability that it will land on the same side each time?

Coin is tossed 3 times => Total number of cases = \(2^3\) = 8

Lets solve the problem using two methods

Method 1:

Out of the 8 cases there are only two cases in which all the outcomes are same. HHH and TTT.

=> Probability that all three produce the same result = \(\frac{2}{8}\) = \(\frac{1}{4}\)

So, Answer will be B

Method 2:

Let's find out the number of cases.
In the first toss we can get anything out of head or tail in 2 ways
In second toss we need to get the same value (head or tail) as we got in the first toss => 1 way
In third toss we need to get the same value (head or tail) as we got in the second toss => 1 way

=> Total number of ways = 2*1*1 = 2 ways

=> Probability that all three produce the same result = \(\frac{2}{8}\) = \(\frac{1}{4}\)

So, Answer will be B
Hope it helps!

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


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