Why do you think this method acting is used ? This is because the hypothesis of obtaining a Head in a coin convulse is angstrom likely as obtaining a tail, that is, 50 %. then when you toss one mint, there are only two possibilities – a steer ( H ) or a dock ( 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 the coin discard probability rule .

**Coin Toss Probability**

probability is the measurement of chances – the 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. then if an event is improbable to occur, its probability is 0. And 1 indicates the certainty for the happening .

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

Number of possible outcomes = 2

Number of outcomes to get drumhead = 1

probability of getting a heading = ½

therefore ,

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

We can generalise the coin flip probability rule :

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

### Solved Examples

**Question** : Two average coins are tossed simultaneously. What is the probability of getting alone one steer ?

**Solution** :

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

thus, the entire act of possible outcomes = 4

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

So number of desire outcomes = 2

therefore, probability of getting only one head

\(\begin{array}{l}=\frac{Number\;of\;favourable\;outcomes}{Total\;number\;of\;possible\;outcomes}\end{array} \)

\(\begin{array}{l}=\frac{2}{4}=\frac{1}{2}\end{array} \)

**Question** : Three fairly 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, full issue of possible outcomes = 8

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

So issue of desire outcomes = 4

therefore, probability of getting at least 2 tails =

\(\begin{array}{l}\frac{No\;of\;favourable\;outcomes}{Total\;number\;of\;possible\;outcomes}\end{array} \)

Read more: How to send your Coin Master link?

\(\begin{array}{l}=\frac{4}{8}=\frac{1}{2}\end{array} \)

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