A coin is tossed 4 times. What is the probability that the

Question

A coin is tossed 4 times. What is the probability that the number of Heads is equal to the number of Tails?

A. 1/8
B. 1/4
C. 3/8
D. 1/2
E. 9/16

help!

GyanOne · Answer

We have to find the probability that we get 2 heads and 2 tails in 4 tosses of the coin.

Using the binary formula and defining a head as a success, 
P(2 heads in 4 tosses) = 4C2* (1/2)^2 * (1/2)^2
 = 6/16
 = 3/8

The answer is therefore (C)

Bunuel · Answer

Solution: We need the probability of: HHTT. P(HHTT)=\frac{4!}{2!*2!}*(\frac{1}{2})^2*(\frac{1}{2})^2=\frac{3}{8} . Explanation: \frac{4!}{2!*2!}=6 is number of all possible cases for our wining scenarios HHTT to appear (which is number of permutation of 4 letters HHTT where 2 H's and 2T's are identical: HHTT, HTHT, HTTH, TTHH, THTH, THHT), (\frac{1}{2})^2 is the probability of HH and (\frac{1}{2})^2 is the probability of TT. Theory behind the concept above: Permutations of n things of which P_1 are alike of one kind, P_2 are alike of second kind, P_3 are alike of third kind ... P_r are alike of r_{th} kind such that: P_1+P_2+P_3+..+P_r=n is: \frac{n!}{P_1!*P_2!*P_3!*...*P_r!} . For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word. Number of permutation of the letters of the word "google" is \frac{6!}{2!2!} , as there are 6 letters out of which "g" and "o" are represented twice. Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \frac{9!}{4!3!2!} . General theory on this topic: Probability: math-probability-87244.html Combinatorics: math-combinatorics-87345.html Hope it helps.

metallicafan · Answer

+1 C

First, we have to calculate the total ways of organizing HHTT. (H: Head; T: Tail):
 \frac{4!}{2!2!} = 6

In any of these ways, the probability is:

\frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} = \frac{1}{16}

So, we multiply that probability by the number of ways (6):

\frac{1}{16} * 6 = \frac{6}{16} = \frac{3}{8}

Answer: C

manishk30 · Answer

Heads equal to tails would be : 2 heads and 2 tails.
i.e. HHTT
4!/2! 2! = 6
6/ 2^4 = 3/8
Ans. C

BrushMyQuant · Answer

Given that a fair coin is tossed 4 times, and we need to find What is the probability that the number of Heads is equal to the number of Tails?

Coin is flipped 4 times => Total number of cases =  2^4  = 16

Number of heads = Number of tails =  \frac{4}{2}  = 2 each

To find Probability of 2H and 2T, it is enough to find the probability of 2H (As other outcomes will be tails then)

To find P(2H) we need to find the two places out of 4 _ _ _ _ in which we will get heads. This can be done in 4C2 ways

=>  \frac{4!}{2!*(4-2)!}  =  \frac{4*3*2!}{2!*2!}  = 6 ways

=> P(2H) =  \frac{6}{16}  =  \frac{3}{8}

So,  Answer will be C 
Hope it helps!

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

https://www.youtube.com/watch?v=LRURv38kZw8

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