A coin is tossed 6 times in succession find the probability of obtaining one head

Probability is a part of mathematics that deals with the possibility of happening of events. It is to forecast that what are the possible chances that the events will occur or the event will not occur. The probability as a number lies between 0 and 1 only and can also be written in the form of a percentage or fraction. The probability of likely event A is often written as P(A). Here P shows the possibility and A shows the happening of an event. Similarly, the probability of any event is often written as P(). When the end outcome of an event is not confirmed we use the probabilities of certain outcomes—how likely they occur or what are the chances of their occurring.

To understand probability more accurately we take an example as rolling a dice:

The possible outcomes are — 1, 2, 3, 4, 5, and 6.

The probability of getting any of the outcomes is 1/6. As the possibility of happening of an event is an equally likely event so there are same chances of getting any number in this case it is either 1/6 or 50/3%.

Formula of Probability

Probability of an event, P(A) = (Number of ways it can occur) ⁄ (Total number of outcomes)

Types of Events

• Equally Likely Events: After rolling dice, the probability of getting any of the likely events is 1/6. As the event is an equally likely event so there is same possibility of getting any number in this case it is either 1/6 in fair dice rolling.
• Complementary Events: There is a possibility of only two outcomes which is an event will occur or not. Like a person will play or not play, buying a laptop or not buying a laptop, etc. are examples of complementary events.

If a coin is flipped 7 times, then what is the probability of getting 6 heads?

Solution:

Use the binomial distribution directly. Let us assume that the number of heads is represented by x  (where a result of heads is regarded as success) and in this case X = 6

Assuming that the coin is unbiased, you have a probability of success ‘p’(where p is considered as success) is 1/2 and the probability of failure ‘q’ is 1/2(where q is considered as failure). The number of trials is represented by the letter ’n’ and for this question n = 7.

Now just use the probability function for a binomial distribution:

P(X = x) = nCxpxqn-x

Using the information in the problem we get

P(X = 6) = (7C6)(1/2)6(1/2)1 

= 7 × 1/64 × 1/2

= 7/128

Hence, the probability of flipping a coin 7 times and getting heads 6 times is 7/128.

Similar Questions

Question 1: What is the probability of flipping a coin 20 times and getting 10 heads?

Answer:

Each coin can either land on heads or on tails, 2 choices.  

(According to the binomial concept)

This gives us a total of 220 possibilities for flipping 20 coins.

Now how many ways can we get 10 heads? This is 20 choose 10, or (20C10)  

This means our probability is (20C10)/220= 184756⁄1048576 ≈ .1762

Question 2: What is the probability of 6 heads in 6 coins tossed together.?

Solution:

6 coin tosses. This means,

Total observations = 36(According to binomial concept)  

Required outcome → 6 Heads {H,H,H,H,H,H}

This can occur only ONCE!

Thus, required outcome = 1  

Probability (6 Heads) = (1⁄2)6 = 1/64

Solution:

Given, a coin is tossed 3 times.

We have to find the probability of getting at least one head.

The possible outcomes are

T H H

H T H

H H T

T T H

T H T

H T T

T T T

H H H

We observe that there is only one scenario in throwing all coins where there are no heads.

The chances for one given coin to be heads is 1/2.

The chance for all three to have the same result would be (1/2)3

= (1/2)(1/2)(1/2)

= 1/8

So, the probability to have atleast one head is 1 - 1/8

= (8 - 1)/8

= 7/8

Therefore, the probability of getting at least one head is 7/8.

A coin is tossed 3 times. What is the probability of getting at least one head?

Summary:

A coin is tossed 3 times. The probability of getting at least one head is 7/8.

Age 11 to 14
Challenge Level

We received a large number of responses of excellent quality.

  Ben from St Peter's followed the tree diagram and calculated out the answer:

If you flip a coin three times the chance of getting at least one head is 87.5%. To get this outcome I used the provided tree diagram to establish how many outcomes used one head.

Llewellyn from St Peter's and Diamor from Willington County Grammar School both observed an interesting pattern and expanded the answer to flipping ten coins:

If you flip a coin 3 times the probability of getting at least one heads is 7 in 8 by reading the table. This table also works the opposite way, the chances of Charlie getting no heads is 1 in 8 because out of all the outcomes only one of them has only tails. I notice that if you add these probabilities together you get the total amount of outcomes (7+1=8). If you flip a coin 4 times the probability of you getting at least one heads is 15 in 16 because you times the amount of outcomes you can get by flipping 3 coins by 2, it results in 16 and then you minus 1 from it. With 5 coins to flip you just times 16 by 2 and then minus 1, so it would result with a 31 in 32 chance of getting at least one heads. With 6 coins you times by 2 and minus by 1 again resulting in a 63 in 64 chance. To find the chance of getting at least one heads if you flip ten coins you times 64 by 2 four times or by 16 once and then minus 1, this results in a 1063 in 1064 chance of getting at least one heads.

  Neeraj from Wilson School developed a generalization for different numbers of possible outcomes:

I noticed that when you add the probabilities together they make a whole. A quick way of figuring out how many times you get at least one head is, that it is always the no. Of possible outcomes minus one over the no. of possible outcomes So: if No of possible outcomes = n the equation would be: P= (n- 1)/n

  One student suggested how to calculate the number of desired outcomes:

If the number of times flipped =p Then the number of outcomes that contain a head is$2^p-1$
So for flipping a coin $10$ times, the number of outcomes with at least one head is $2^{10}-1 = 1024 - 1 = 1023$

  Luke from Maidstone Grammar School went further to investigate the next part of the question:

When there are 4 green balls in the bag and there are 6 red balls the probability of randomly selecting a green ball is 0.4 ($\frac{2}{5}$) and the probability selecting a red ball is 0.6 ($\frac{3}{5}$).

If a ball is selected and then replaced the probability of picking a red ball or a green ball is the same every time. When 3 balls are picked with replacement the probability of getting at least one green is
1-(the probability of getting 3 reds)
Because the probability is the same every time the chance of getting 3 reds is $0.6^3=0.216$ (or in fractions $(\frac{3}{5})^3 = \frac{27}{125}$). So the probability of getting at least one green is $1-0.216=0.784$ (or in fractions $1 - \frac{27}{125} = \frac{98}{125}$).

  When the balls are not replaced the probability of getting at least one green is still 1-(the probability of getting 3 reds). In each draw the probability of drawing a red ball is $\frac{\text{the number of red balls}}{\text{the total number of balls}}$
On the first draw there are 6 red balls out of 10 so the probability of picking a red is $\frac{6}{10}$.
On the second draw there are 5 red balls out of 9 so the probability of picking a red is $\frac{5}{9}$.
On the final draw there are 4 red balls out of 8 so the probability of picking a red is $\frac{4}{8}$.
The probability of this sequence of draws happening is the probability of each draw multiplied together. i.e.: $\frac{6}{10}\times\frac{5}{9}\times\frac{4}{8}=\frac{1}{6}$
The probability of drawing all reds is $\frac{1}{6}$ and so the probability of drawing at least one green is $\frac{5}{6}$.

  Helen from Stroud finished up the problem:

  When children are selected for the school council they are not replaced. The children are selected one after another and each time the probability of a boy being selected is
P(boy selected first) = $\frac{\text{the number of boys in the class}}{\text{the total number of children in the class}}$
Note: the class refers to students who have not already been made part of the council.
To find the probability that there will be at least one boy, find the probability that all three are girls, and then
P(at least one boy selected) = 1-P(all girls selected)
to get the answer.
The probability of picking a girl is
P(girl selected first) = $\frac{\text{number of girls in class}}{\text{total number in class}}= \frac{15}{28}$
Then P(second selected also a girl) = $\frac{14}{27}$
And P(third selected also a girl) = $\frac{13}{26}$
So P(all girls selected) = $\frac{15}{28}\times\frac{14}{27}\times\frac{13}{26} = \frac{5}{36}$
Then the answer is
P(at least one boy selected) = 1 - P(all girls selected) = 1 - $\frac{5}{36}$ = $\frac{31}{36}$

  Well done to everyone.

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