A fair coin is tossed 6 times. What is the probability that exactly 2

Question

A fair coin is tossed 6 times. What is the probability that exactly 2 heads will show.

A. 1/64
B. 7/64
C. 9/64
D. 15/64
E. 49/64

firas92 · Answer

We need 2 heads and 4 tails : HHTTTT

No of ways this can be achieved is  \frac{6!}{4!2!} = 15

In each case, the probability is  (\frac{1}{2})^2(\frac{1}{2})^4 = \frac{1}{64}

So, probability that exactly 2 heads will show  = \frac{15}{64}

Answer is  (D)

GMATWhizTeam · Answer

Solution: 
The coin is tossed exactly 6 times.
 • On each flip, there are only two possible outcomes, i.e., either a head or a tail.
 o The coin is fair so the probability of getting head or tail is equal on each toss.  
 • Number of outcomes on 6 tosses = 2*2*2*2*2*2 = 64 
The number of ways in which 2 heads will show on 6 tosses = 6C2 =  \frac{6!}{4!*2!}  = 15
 • Probability = \frac{ 15}{64}  
Hence, the correct answer is Option D.

sujoykrdatta · Answer

We can treat this as an ARRANGEMENT or a SELECTION problem. Let us see how:

Approach 1:

We need to have  H H T T T T  (2 Heads and 4 Tails)

This can be treated as an ARRANGEMENT Problem:

Number of ways of arranging 2 H and 4 T =  \frac{6!}{(2! * 4!)}  =  \frac{(6 * 5 * 4!)}{(2 * 4!)}  = 15

Total number of ways in which 6 coins can be tossed (where each coin has 2 options - either Head or Tail) =  2^6 = 64

Required probability =  \frac{15}{64}

Answer D

Approach 2:

We need 2 Heads (and hence, 4 Tails)

We can treat this as a SELECTION problem where we need to select 2 Heads from 6 coins

Number of ways of selecting =  6C2  =  \frac{6!}{(2! * 4!)}  = 15

Total number of ways in which 6 coins can be tossed (where each coin has 2 options - either Head or Tail) =  2^6 = 64

Required probability =  \frac{15}{64}

Answer D

Jawad001 · Answer

6C2 *(1/2)^2*(1/2)^4 = 15*(1/64) = 15/64

BrushMyQuant · Answer

Given that A fair coin is tossed 6 times and we need to find What is the probability that exactly 2 heads will show.

Coin is tossed 6 times => Total number of cases =  2^6  = 64

Cases in which we will get 2 Heads

Cases in which we get 2H can be found by putting 2 heads in any of the 6 slots _ _ _ _ _ _ _ 
This can be done in 6C2 ways =  \frac{6!}{2!*(6-2)!}  =  \frac{6*5*4!}{2!*4!}  = 15

=> P(6H) =  \frac{15}{64}

So,  Answer will be D 
Hope it helps!

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

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

bumpbot · Answer

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A fair coin is tossed 6 times. What is the probability that exactly 2

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A fair coin is tossed 6 times. What is the probability that exactly 2

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