Chance of Two Heads

All possible ways to dispose two elements (heads or tails) in six positions (coin tosses) are 2^6 = 64. All possible ways to dispose two specific elements (two heads) in six position are 6 choose 2 {6 \choose 2} = 15. Therefore the probability of getting two exactly heads with six coin tosses is \frac {15}{64}. 15 and 64 are already coprime, then the answer is 15 + 64 = 79.

Total outcomes when tossing a coin 1 time = 2

Total outcomes when tossing a coin 6 times = 2^6 = 64

We want two heads, simply we want the anagram TTTTHH.

Combination = 6!/4!2! = 720/48 = 15.

So the probability is 15/64 and a+b = 15+64 = 79.

the coin has been flipped 6 times , so there are 2^6 possible outcomes.the favorable outcomes contain exactly 2 heads and the order in which they can occur can be anything. so, the number of ways in which 2 heads can appear out the 6 possible flips is 6c2.this gives p(e) = 6c2/2^6.

Total combination per coin flip = 2.

Here there are 6 flips, so total combination = 2^6 = 64.

We want the combination of anagram HHTTTT, which is equal to 6!/2!4! = 720/2*24 = 720/48 = 15

Thus the probablity is 15/64 and the value of 15+64 = 79

There are $2^6 = 64$ different outcomes.

Let the coins be in a line, and label them $a_1, a_2, ..., a_6$ from first to last. Exactly $2$ of them would be heads, thus we could "choose" $2$ of them to be heads. Since there are $6$ elements in the set $a_1, a_2, ..., a_6$ , there are $6C2 = 15$ ways.

$a= 15$ , $b=64$ , thus, $a+b=\boxed{79}$

a has to be the number of needed cases and b the number of total cases. The number of wanted cases is equal to the number of possible way of arranging 2 X and 4 Y, which is 6!/(2!*4!)= 15. The number of total cases is simply 2^6=64. a/b=15/64, therefore a+b=15+64=79

The probability of getting any combination of heads and tails in six flips is (1/2)^6 = 1/64. Multiply that probability by the number of combinations of 2 heads and 4 tails: 6C2 = 6!/(2!4!) = 15.

The probability, or a/b, is 15/64. Therefore, a + b = 79.

We can think of every 6 flips of the coin as a set. Therefore, there will be $2^6 = 64$ possible distinct sets, that is, 64 different ways the 6 flips of the coin could turn out.

But, out of these 6 flips, we want exactly 2 flips to show heads, so the number of desired sets would be ${6 \choose {2}} = 15$ .

Thus, the probability of getting 2 heads is $\frac {a}{b} = \frac {15}{64}$ . So, $a+b = 15+64 = \boxed {79}$

We use Binomial Probability. There are $\dbinom {6}{4}$ ways of getting two heads and four tails in a series of six tosses. There is a total of $2^6 = 64$ possible toss series results. Thus, the probability is $\dfrac {\dbinom {6}{4} = 15}{64} \implies a + b = 15 + 64 = \boxed {79}.$

1. Find the number of combinations which include exactly getting 2 heads. -> binomial(6,2) = 15.
2. Divide it by the total number of combinations possible. This is 2^{6} = 64.
3. The probability is \frac{15}{64}. The sum of a+b is 15 + 64 = 79

The main technique we are using now is called listing . More advanced, alternative techniques include Pascal's Triangle or the binomial coefficient .

Let us first consider the trivial case of flipping one coin. There are two distinct possibilities: H or T. It is quite obvious to see that there is a 50% chance of getting heads or tails ( tails means no heads ).

Now let's move on to flipping 2 coins. The possible combinations are as follows:

1. TT
2. T H
3. H T
4. HH

We can see here that there is a one-in four chance of 0 heads, two-in four chance of 1 head, and a one-in-four chance of 2 heads. Formatted nicely, it is as follows:

$\frac{1}{2} , \frac{1}{2}$
$\frac{1}{4} , \frac{1}{2}, \frac{1}{4}$

Can we do this for three coins? Writing the possible combinations once more:

1. TTT
2. H TT
3. T H T
4. TT H
5. HH T
6. H T H
7. T HH
8. HHH

There are eight possible combinations. But if you are new to listing, please take note that it is important to list in a systematic manner! I have bolded each H so it stands out more. Notice that I start with one H at the first position, and systematically move it forward? After it moves to the end I then repeat the process with two Hs. This is very important! You may miss possibilities out otherwise. Anyway, what we have now is as follows:

$\frac{1}{2} , \frac{1}{2}$
$\frac{1}{4} , \frac{1}{2}, \frac{1}{4}$
$\frac{1}{8} , \frac{3}{8}, \frac{3}{8}, \frac{1}{8}$

Is there a pattern here? There are three things we seem to be getting from this:

1. The sum of every possible combination adds up to 1. This makes intuitive sense, since adding up these fractions is the same as adding up percentages, and the sum of probabilities will always equal to a 100% - that is, a sure thing.

2. The number of distinct possibilities seems to be equal to $2^{n}$ , where ${n}$ is the number of coins flipped. This seems confusing at first! But it actually makes sense. As each coin has two outcomes, heads or tails, the number of total possible outcomes is simply equal to the product of the possible outcomes of each individual coin ! That is, $2 \times 2 \times 2 \times 2....$ in this case! (Extension: how many possible outcomes are there with a coin and a six-sided die?)

3. This pattern resembles Pascal's Triangle . Pascal's Triangle is a triangle constructed as follows: The value of a number is given by the sum of the two numbers above it. Pascal's Triangle is also related to the binomial coefficient . If you know these concepts already, then you can simply look up the second number of the 6th row of Pascal's Triangle, or 6 choose 2. But if you do not know it, it is also OK. You can simply list the sixty-four combinations.

Luckily, we don't actually need to list all sixty-four. We can just start listing the front few.

No heads

1. TTTTTT

*One head *

1. HTTTTT
2. THTTTT
3. TTHTTT
4. TTTHTT
5. TTTTHT
6. TTTTTH

Two heads

1. HH TTTT
2. H T H TTT
3. H TT H TT
4. H TTT H T
5. H TTTT H
6. T HH TTT
7. T H T H TT
8. T H TT H T
9. T H TTT H
10. TT HH TT
11. TT H T H T
12. TT H TT H
13. TTT HH T
14. TTT H T H
15. TTTT HH

There we have it! There are fifteen possible combinations out of a possible sixty-four , leaving us with the fraction of $\frac{15}{64}$

Since the numerator and the denominator are coprime, the sum of these two numbers is 79, which gives us the correct answer.

According to Bernoulli trial,

Probability of x successful trials out of n trials is given by:

$p(x) =\binom{n}{x}p^{x}q^{n-x}$

where $p$ and $q$ are the probabilities of success and failure in each trial

For a coin toss, there are 2 results in each toss i.e head(success) and tails(failure)

Hence, $p = \frac{1}{2} , q=\frac{1}{2}$

Hence, probability of getting 2 heads in 6 toss = $\binom{6}{2}(\frac{1}{2})^{2}(\frac{1}{2})^{4}$

= $\frac{15}{64}$

Hence, $a + b = 15 + 64 = 79$

probability of getting head is 1/2 (successful outcome). nCr P^{n-r}Q^{r} P -> successful outcome Q -> unsuccessful outcome so here it is 6C2 (1/2)^{4}*(1/2)^{2} =15/64 so a+b=15+64 =79

We observe that number of cases to get exactly two heads are 6! / (4!*2!) as the event {H,H,T,T,T,T} has 4 identical objects namely event tail and two identical other object namely head,

And total number of cases are 2^6 as each position of the 6 tosses has two options namely head and tail.

Therefore probability is (6!/(4!*2!))/(2^6) and we get a = 15 and b = 64 hence a+b = 79

a= 6C2 , B=2^6 , a+b=79

Suppose head with H, and Complement with G

$P(HHGGGG)=\frac{1}{2^6}$

Suppose, probability we getting excactly 2 heads are $P(A)$

So, $PA=C(6,2) \times P(HHGGGG)=\frac{15}{64}$

$a+b=15+64=\fbox{89}$ as desired.

Consider the outcome as a binary string of length 6, where 1 is heads and 0 is tails. There are $2^6 = 64$ possibilities in total, out of those we want the number of strings where there are only two 1's, so we do $\binom{6}{2} = 15$ . So the probability is $\frac{15}{64}$ . $15 + 64 = 79$ .

The total number of possible permutations are $2^{6} = 64$

The feasible outcomes are ${{6}\choose{2}} = 15$ because exactly two out of the six spots have to be occupied by heads and the order of the heads doesn't matter, that is, $\left(h_1,h_2\right)$ is the same as $\left(h_2,h_1\right)$ .

The probability is $\dfrac{15}{64}$ , and so the answer is $\boxed{79}$

n(S)=2^6=64, n(A)=6!/(2!x4!)=15, P(A)=15/64. So a+b = 15+64 = 79

There are 6C2=15 ways for there to be exactly 2 heads out of the 64 possible out comes. 15 and 64 are coprime so the answer is 15+64=79

This is an example of Bernoulli's Trials (or Distributions). Let us choose the 2 trials when the heads occurs. This can be chosen in 6C2 which is equal to 15 ways. Automatically the other 4 trials will be where tails occur. Now the probability of getting heads in the two trials chosen by us is 1/2*1/2 ( By the product rule in Probability where we multiply if a AND case is present in the two events). Now similarly the probability of getting 4 tails is 1/2 *1/2 * 1/2 * 1/2 (again by product rule). Hence in all the probability of getting 2 heads and 4 tails is :

6C2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2

= 15 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2

= 15/64

Hence a=15 b=64 and a+b =79

Voila !

• n(E); There're 6 coins, choose 2 to be heads. (C(6,2) ways.)
• n(S); There're 6 coins, each coins has 2 ways to show (H or T). (2^6 ways.)
• Therefore. P(E) = C(6,2)/2^6 = 15/64 = a/b -> a+b = 79

Assume that the flipped function f(x,y)=(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

If x=head then x^2 as 2 heads , we get 15. The sum of whole coefficient is 64. Since a=15 and b=64. Then a+b=79

Assume that the flipped function f(x,y)=(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

If x=head then x^2 as 2 heads , we get 15. The sum of whole coefficient is 64. Since a=15 and b=64. Then a+b=79

Use the knowledge of permutation combination combined with bernoulis probability theorem Multiply 1/2 six times for the events and select any 2 out of six chances to get 5 he desired result

Coin flipped =6 times, probability of getting head=2,and can be written as a/b . a,b are co-prime numbers and positive integers ,so from this given clue.....we can find a+b.

its easy firstly, there are 2^6 possibilities,there are 6C2 kinds of choices therefore, 6C2/2^6 = 15/64

All cases 2^6=64

get only tow heads 6!/(4!*2!)=15 The probability a/b=15/64
a+b=79