What is the probability of obtaining two tails and one head when three coins are tossed?

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The probability of some event happening is a mathematical (numerical) representation of how likely it is to happen, where a probability of 1 means that an event will always happen, while a probability of 0 means that it will never happen. Classical probability problems often need to you find how often one outcome occurs versus another, and how one event happening affects the probability of future events happening. When you look at all the things that may occur, the formula (just as our coin flip probability formula) states that

probability = (no. of successful results) / (no. of all possible results).

Take a die roll as an example. If you have a standard, 6-face die, then there are six possible outcomes, namely the numbers from 1 to 6. If it is a fair die, then the likelihood of each of these results is the same, i.e., 1 in 6 or 1 / 6. Therefore, the probability of obtaining 6 when you roll the die is 1 / 6. The probability is the same for 3. Or 2. You get the drill. If you don't believe me, take a dice and roll it a few times and note the results. Remember that the more times you repeat an experiment, the more trustworthy the results. So go on, roll it, say, a thousand times. We'll be waiting here until you get back to tell us we've been right all along. Go to the dice probability calculator if you want a shortcut.

But what if you repeat an experiment a hundred times and want to find the odds that you'll obtain a fixed result at least 20 times?

Let's look at another example. Say that you're a teenager straight out of middle school and decide that you want to meet the love of your life this year. More specifically, you want to ask ten girls out and go on a date with only four of them. One of those has got to be the one, right? The first thing you have to do in this situation is look in the mirror and rate how likely a girl is to agree to go out with you when you start talking to her. If you have problems with assessing your looks fairly, go downstairs and let your grandma tell you what a handsome, young gentleman you are. So a solid 9 / 10 then.

As you only want to go on four dates, that means you only want four of your romance attempts to succeed. This has an outcome of 9 / 10. This means that you want the other six girls to reject you, which, based on your good looks, has only a 1 / 10 change of happening (The sum of all events happening is always equal to 1, so we get this number by subtracting 9 / 10 from 1). If you multiply the probability of each event by itself the number of times you want it to occur, you get the chance that your scenario will come true. In this case, your odds are 210 * (9 / 10)4 * (1 / 10)6 = 0.000137781, where the 210 comes from the number of possible fours of girls among the ten that would agree. Not very likely to happen, is it? Maybe you should try being less beautiful!

Here we will learn how to find the probability of tossing three coins.

Let us take the experiment of tossing three coins simultaneously:

When we toss three coins simultaneously then the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for head and T is denoted for tail.

Therefore, total numbers of outcome are 23 = 8

The above explanation will help us to solve the problems on finding the probability of tossing three coins.

Worked-out problems on probability involving tossing or throwing or flipping three coins:

1. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times.

If three coins are tossed simultaneously at random, find the probability of:

(i) getting three heads,

(ii) getting two heads,

(iii) getting one head,

(iv) getting no head

Solution:

Total number of trials = 250.

Number of times three heads appeared = 70.

Number of times two heads appeared = 55.

Number of times one head appeared = 75.

Number of times no head appeared = 50.

In a random toss of 3 coins, let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and 0 head respectively. Then,

(i) getting three heads

P(getting three heads) = P(E1) Number of times three heads appeared

= Total number of trials

= 70/250

= 0.28

(ii) getting two heads

P(getting two heads) = P(E2) Number of times two heads appeared

= Total number of trials

= 55/250

= 0.22

(iii) getting one head

P(getting one head) = P(E3) Number of times one head appeared

= Total number of trials

= 75/250

= 0.30

(iv) getting no head

P(getting no head) = P(E4) Number of times on head appeared

= Total number of trials

= 50/250

= 0.20

Note:

In tossing 3 coins simultaneously, the only possible outcomes are E1, E2, E3, E4 and P(E1) + P(E2) + P(E3) + P(E4)

= (0.28 + 0.22 + 0.30 + 0.20)

= 1

2. When 3 unbiased coins are tossed once.

What is the probability of:

(i) getting all heads

(ii) getting two heads

(iii) getting one head

(iv) getting at least 1 head

(v) getting at least 2 heads

(vi) getting atmost 2 heads

Solution:

In tossing three coins, the sample space is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

And, therefore, n(S) = 8.

(i) getting all heads

Let E1 = event of getting all heads. Then,
E1 = {HHH}
and, therefore, n(E1) = 1.
Therefore, P(getting all heads) = P(E1) = n(E1)/n(S) = 1/8.

(ii) getting two heads

Let E2 = event of getting 2 heads. Then,
E2 = {HHT, HTH, THH}
and, therefore, n(E2) = 3.
Therefore, P(getting 2 heads) = P(E2) = n(E2)/n(S) = 3/8.

(iii) getting one head

Let E3 = event of getting 1 head. Then,
E3 = {HTT, THT, TTH} and, therefore,
n(E3) = 3.
Therefore, P(getting 1 head) = P(E3) = n(E3)/n(S) = 3/8.

(iv) getting at least 1 head

Let E4 = event of getting at least 1 head. Then,
E4 = {HTT, THT, TTH, HHT, HTH, THH, HHH}
and, therefore, n(E4) = 7.
Therefore, P(getting at least 1 head) = P(E4) = n(E4)/n(S) = 7/8.

(v) getting at least 2 heads

Let E5 = event of getting at least 2 heads. Then,
E5 = {HHT, HTH, THH, HHH}
and, therefore, n(E5) = 4.
Therefore, P(getting at least 2 heads) = P(E5) = n(E5)/n(S) = 4/8 = 1/2.

(vi) getting atmost 2 heads

Let E6 = event of getting atmost 2 heads. Then,
E6 = {HHT, HTH, HTT, THH, THT, TTH, TTT}
and, therefore, n(E6) = 7.
Therefore, P(getting atmost 2 heads) = P(E6) = n(E6)/n(S) = 7/8

3. Three coins are tossed simultaneously 250 times and the outcomes are recorded as given below.

Outcomes	3 heads	2 heads	1 head	No head	Total
Frequencies	48	64	100	38	250

If the three coins are again tossed simultaneously at random, find the probability of getting

(i) 1 head

(ii) 2 heads and 1 tail

(iii) All tails

Solution:

(i) Total number of trials = 250.

Number of times 1 head appears = 100.

Therefore, the probability of getting 1 head

= \(\frac{\textrm{Frequency of Favourable Trials}}{\textrm{Total Number of Trials}}\)

= \(\frac{\textrm{Number of Times 1 Head Appears}}{\textrm{Total Number of Trials}}\)

= \(\frac{100}{250}\)

= \(\frac{2}{5}\)

(ii) Total number of trials = 250.

Number of times 2 heads and 1 tail appears = 64.

[Since, three coins are tossed. So, when there are 2 heads, there will be 1 tail also].

Therefore, the probability of getting 2 heads and 1 tail

= \(\frac{\textrm{Number of Times 2 Heads and 1 Trial appears}}{\textrm{Total Number of Trials}}\)

= \(\frac{64}{250}\)

= \(\frac{32}{125}\)

(iii) Total number of trials = 250.

Number of times all tails appear, that is, no head appears = 38.

Therefore, the probability of getting all tails

= \(\frac{\textrm{Number of Times No Head Appears}}{\textrm{Total Number of Trials}}\)

= \(\frac{38}{250}\)

= \(\frac{19}{125}\).

These examples will help us to solve different types of problems based on probability of tossing three coins.

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