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Bunuel
A fair coin is to be flipped four times. What is the probability that the coin will land on the same side on all four flips?

(A) 1/32
(B) 1/16
(C) 1/8
(D) 1/4
(E) 1/2

Solution



    • We are given that a fair coin is flipped four times.

      o Each time, we will get either a head or a tail.
      o Thus, total possible cases \(= 2 * 2 * 2 * 2 = 16\)
    • We need to find the probability that the coin will land on the same side on all four flips.

      o This is possible only when all the four time, we get all HEADs (HHHH) or all TAILs (TTTT)

      o Thus, total favorable cases \(= 1 + 1 = 2\)

    • Thus, the probability \(= \frac{Favorable Cases}{Total Cases} = \frac{2}{16} = \frac{1}{8}\)

The correct answer is Option C.

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Bunuel
A fair coin is to be flipped four times. What is the probability that the coin will land on the same side on all four flips?

(A) 1/32
(B) 1/16
(C) 1/8
(D) 1/4
(E) 1/2


The probability of all heads is (1/2)^4 = 1/16, and the probability of all tails is 1/16. So, the probability of the coin landing on the same side on all four flips is 2/16 = 1/8.

Alternate solution:

Which side the coin lands the first time doesn’t matter (so the probability is 1), but there is only a ½ chance on each successive flip for the coin to land on the same side as the first time. Therefore, the probability that the coin will land on the same side on all four flips is:

1 x ½ x ½ x ½ = 1/8

Answer: C
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Hi all,

does anyone know how to use the combination formula to solve this question?

= # of ways of selection x (1/2*1/2*1/2*1/2)
= 4C1 * 1/16

but the answer is not correct, can someone please let me know what is incorrect here? thanks much.
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Given that A fair coin is to be flipped four times and We need to find What is the probability that the coin will land on the same side on all four flips?

Coin is tossed 4 times => Total number of cases = \(2^4\) = 16

Coin lands on the same side = Getting 4 Tails or Getting 4 Heads

=> P(4H or 4T) = \(\frac{2}{16}\) (As there is two cases out of 16 where we get 4H or 4T)
= \(\frac{1}{8}\)

So, Answer will be C
Hope it helps!

Playlist on Solved Problems on Probability here

Watch the following video to MASTER Probability with Coin Toss Problems

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So, we have 1/2 chance that it lands on heads and 1/2 chance on tails.

However, since it doesn't specify which side we need, the first flip has a 1/1 probability (or 2/2) that it would either be a heads or a tails.
Then on the second, third and fourth flips we have to get the same as the first flip which is 1/2 as depending on whether we got heads or tails we can only get that side.
So, 1 * 1/2 * 1/2 * 1/2 = 1/2^3 = 1/8
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Bunuel
A fair coin is to be flipped four times. What is the probability that the coin will land on the same side on all four flips?

(A) 1/32
(B) 1/16
(C) 1/8
(D) 1/4
(E) 1/2
All four times may be H or T

H=1/16
T=1/16
H or T= 1/8
­
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You want all four flips to show the same side — meaning either all heads (HHHH) or all tails (TTTT).

Now, the total number of possible outcomes when flipping four coins is 24 = 16.

Only two outcomes meet the condition:

HHHH

TTTT

So, the probability = (Number of favorable outcomes) ÷ (Total outcomes) = 2/16 = 1/8.

If you try this using a Flip 3 coins simulator (flipping three coins first and imagining a fourth flip), you’d notice it’s really rare to get all the same face — that’s why the probability is just 1/8.

Thus, the correct answer is (C) 1/8!

Bunuel
A fair coin is to be flipped four times. What is the probability that the coin will land on the same side on all four flips?

(A) 1/32
(B) 1/16
(C) 1/8
(D) 1/4
(E) 1/2
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