You might have noticed that before the commencement of a cricket match, a decision is to be made as to which team would bat or bowl first. How is this done? You see that the captains of the two teams participate in a coin toss wherein they pick one side of a coin each – say head or tail. The umpire tosses the coin in the air. The team which wins the toss gets to make the decision of batting or bowling first. This is one of the most common applications of the coin toss experiment.

Why do you think this method is used? This is plainly because the possibility of obtaining a head in a coin toss is as likely as obtaining a tail, that is, 50%. So when you toss one coin, there are only two possibilities – a head (H) or a tail (L). However, what if you want to toss 2 coins simultaneously? Or say, 3, 4 or 5 coins? The outcomes of these coin tosses will differ. Let us learn more about coin toss probability formula.

**Coin Toss Probability**

Probability is the measurement of chances – likelihood that an event will occur. If the probability of an event is high, it is more likely that the event will happen. It is measured between 0 and 1, inclusive. So if an event is unlikely to occur, its probability is 0. And 1 indicates the certainty for the occurrence.

Now if I ask you what is the probability of getting a head when you toss a coin? Assuming the coin to be fair, you straightaway answer 50% or ½. This is because you know that the outcome will either be head or tail, and both are equally likely. So we can conclude here:

Number of possible outcomes = 2

Number of outcomes to get head = 1

Probability of getting a head = ½

Hence,

\(Probability\;of\;getting\;a\;head = \frac{No\;of\;outcomes\;to get\;head}{No\;of\;possible\;outcomes}\)

We can generalise the coin toss probability formula:

\(Probability\;of\;certain\;event=\frac{Number\;of\;favourable\;outcomes}{Total\;number\;of\;possible\;outcomes}\)

**Solved Examples**

**Question**: Two fair coins are tossed simultaneously. What is the probability of getting only one head?

**Solution**:

When 2 coins are tossed, the possible outcomes can be {HH, TT, HT, TH}.

Thus, total number of possible outcomes = 4

Getting only one head includes {HT, TH} outcomes.

So number of desired outcomes = 2

Therefore, probability of getting only one head

\(=\frac{Number\;of\;favourable\;outcomes}{Total\;number\;of\;possible\;outcomes}\)

\(=\frac{2}{4}=\frac{1}{2}\)

**Question**: Three fair coins are tossed simultaneously. What is the probability of getting at least 2 tails?

**Solution**:

When 3 coins are tossed, the possible outcomes can be {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

Thus, total number of possible outcomes = 8

Getting at least 2 tails includes {HTT, THT, TTH, TTT} outcomes.

So number of desired outcomes = 4

Therefore, probability of getting at least 2 tails =

\(\frac{No\;of\;favourable\;outcomes}{Total\;number\;of\;possible\;outcomes}\)

\(=\frac{4}{8}=\frac{1}{2}\)

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