A fair coin is tossed six times. What is the probability that the resu

Question

A fair coin is tossed six times. What is the probability that the result will be exactly three heads and three tails?

A. 1/6

B. 1/4

C. 5/16

D. 1/2

E. 13/24

aritrar4 · Answer

Formula for exact occurrence =  C^N_kP(A)^kP(B)^{N-k}

Heads = P(A) = 1/2
Tails = P(B) = 1/2

So required probability =  C^6_3\frac{1}{2}^3\frac{1}{2}^{6-3}

C^6_3*\frac{1}{2}^6

Ans 5/16 C IMO

Mona2019 · Answer

Probability of getting 3 heads and 3 tails is (1/2)^3*(1/2)^3
As probability of getting a head =1/2= probability of getting a tail
Total ways we can arrange 3heads and 3 tails is 6!/(3!*3!)=5/16

IMO c is the answer

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Archit3110 · Answer

3 tails and 3 heads arrangement ; 6!/3!*3! ; 20
total tosses ; (1/2)^6 ; 1/64

20*/64 ; 10/32 ; 5/16
option C

Kinshook · Answer

Given: A fair coin is tossed six times. 
Asked: What is the probability that the result will be exactly three heads and three tails?

Probability that the result will be exactly three heads and three tails =  ^6C_3 * (\frac{1}{2})^3 * (\frac{1}{2})^{(6-3)} = \frac{20}{2^6} = \frac{5}{2^4} = \frac{5}{16}

IMO C

hellothere88 · Answer

Can you explain how is formula of probability of exact occurance derived?

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GmatKnightTutor · Answer

It might be helpful to understand the underlying concept, people.

ScottTargetTestPrep · Answer

Solution:

The probability that the result will be exactly three heads and three tails, in that exact order, is:

½ x ½ x ½ x ½ x ½ x ½ = 1/64

However, since three heads and three tails can be arranged in 6C3 = 6! / (3!3!) = 720/36 = 20 ways, the probability that the result will be exactly three heads and three tails (in any order) is:
1/64 x 20 = 20/64 = 5/16

Answer: C

BrushMyQuant · Answer

We are given that A fair coin is tossed six times and we need to find What is the probability that the result will be exactly three heads and three tails

Now there are six places to fill as shown below

_ _ _ _ _ _

We need to get 3 Heads and 3 Tails.
Now lets find out the slots out of these 6 in which 3 heads will go.

We can find that using 6C3 =  \frac{6!}{3!*(6-3)!}  =  \frac{6!}{3!*3!}  =  \frac{6*5*4*3*2*1}{3*2*1*3*2*1}  = 20 ways

Now, in the remaining slots we will have Tails. So we can get 3H and 3T in 20 ways

WE know that probability of getting a head, P(H), = Probability of getting a Tail, P(T) =  \frac{1}{2}

=> Probability of getting 3H and 3T = Number of ways * P(H) * P(H) * P(H) * P(T) * P(T) * P(T) = 20 *  \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2}  =  \frac{5}{16}

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

egmat · Answer

The Formula Has Two Parts: 
 Probability of exact k successes = C(N,k) × P(success)^k × P(failure)^(N-k)

Part 1: Probability of ONE specific sequence 
Let's say we want EXACTLY the sequence: H H H T T T (heads first, then tails)

Since each coin toss is  independent :
P(H H H T T T) = (1/2) × (1/2) × (1/2) × (1/2) × (1/2) × (1/2) = (1/2)^6

In general, for ANY specific sequence with k heads and (N-k) tails:
P(one specific sequence) =  P(H)^k × P(T)^(N-k)

Part 2: Count HOW MANY such sequences exist 
But wait. H H H T T T is not the ONLY way to get 3 heads and 3 tails!

We could also get:
- H T H T H T
- T H T H T H
- H H T T H T
- ...and many more!

How many arrangements give us exactly 3 heads among 6 tosses?

This is like asking: "In how many ways can we choose  3 positions  (out of 6) to place the heads?"
Answer: C(6,3) =  20  different sequences

Putting It Together: 
- Each of these  20  sequences has the same probability: (1/2)^6
- These sequences are  mutually exclusive  (can't happen at the same time)
- So we ADD their probabilities: 20 × (1/2)^6 = 20/64 =  5/16

General Principle:

The formula = (Number of favorable arrangements) × (Probability of each arrangement)
 = C(N,k) × P(success)^k × P(failure)^(N-k)

Simple Example: 
Flip a coin  3  times. Probability of exactly  2  heads?

Step 1: One specific sequence like HHT has probability = (1/2)^3 = 1/8
Step 2: How many ways to arrange 2 H's in 3 spots? C(3,2) =  3  ways (HHT, HTH, THH)
Step 3: Answer = 3 × (1/8) =  3/8

Answer: C (5/16)

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