When the Euro coin was introduced in $2002$, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of $250$spins,$140$showed a head (event H) while $110$showed a tail (event T). On that basis, they claimed that it is not a fair coin.

a. Based on the given data, find P(H) and P(T).

b. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice.

c. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin.

d. Use the tree to find the probability of obtaining at least one head

We are given, that the Euro coin was introduced in $2002$, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of $250$spins,$140$showed a head (event H) while $110$showed a tail (event T). On that basis, they claimed that it is not a fair coin.

Probability of an event $E$is

$P=\frac{Numberoffavourableoutcomes}{Numberofpossibleoutcomes}$

We have,

Total number of spins$250$

Number of heads$140$

Number of tails $110$

Now the events, H showed the head, T showed the tail,

$$\begin{gathered} P\left(H\right)=\frac{numberofheads}{totalnumberofspins} \\ P\left(H\right)=\frac{140}{250} \\ P\left(T\right)=\frac{numberoftails}{totalnumberofspins} \\ P\left(T\right)=\frac{110}{250} \end{gathered}$$

Probability of an event $E$is

localid="1648140042813" $P=\frac{Numberoffavourableoutcomes}{Numberofpossibleoutcomes}$

The tree presentation of possible outcomes for the experiment of tossing coins twice is

Probability of an event $E$is calculated as follow:

$P=\frac{Numberoffavourableoutcomes}{Numberofpossibleoutcomes}$

The tree presentation of the probability of obtaining exactly one head in two tosses of the coin is shown below

the probability of obtaining exactly one head in two tosses of the coin is calculated as

$$\begin{gathered} P(exactlyonehead)=P\left(HT\right)+P\left(TH\right) \\ P\left(\right(exactlyonehead)=\frac{154}{625}+\frac{154}{625} \\ P((exactlyonehead)=\frac{308}{625} \\ P\left(\right(exactlyonehead)=0.4928 \end{gathered}$$

Hence, the probability of obtaining exactly one head in two tosses of the coin is$0.4928$.

Probability of an event $E$is

$P=\frac{Numberoffavourableoutcomes}{Numberofpossibleoutcomes}$

The tree presentation of the probability of obtaining at least one head is shown below:

The probability of obtaining at least one head is:

$$\begin{gathered} P(atelastonehead)=P\left(HT\right)+P\left(TH\right)+P\left(HH\right) \\ P(atelastonehead)=\frac{154}{250}+\frac{154}{250}+\frac{196}{250} \\ P(atelastonehead)=0.806 \\ P(atelastonehead)=0.81 \end{gathered}$$