Coin Toss Probability formula with Solved Examples

You might have noticed that before the beginning of a cricket catch, a decision is to be made, which team would bat or bowl first. How is this done ? You see that the captains of the two teams participate in a coin convulse wherein they pick one side of a coin each, that is promontory or tail. The referee tosses the coin in the atmosphere. The team which wins the pass gets to make the decisiveness of batting or bowl first. This is one of the most common applications of the coin discard experiment .
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 =

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

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

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source : https://leowiki.com
Category : Economy