Question

Question:In roulette, a wheel with 38 numbers is spun. Of these, 18 are red, and 18 are black. The other two numbers, which are neither black nor red, are 0 and 00. The probability that when the wheel is spun it lands on any particular

number is 1/38.

(a) What is the probability that the wheel lands on a red number.

(b) What is the probability that the wheel lands on a black number twice in a row.

(c) What is the probability that wheel the lands on 0 or 00.

(d) What is the probability that in five spins the wheel never lands on either 0 or 00.

(e) What is the probability that the wheel lands on one of the first six integers on one spin, but does not land on any of them on the next spin.

Short answer

Answer

(a)The probability that the wheel lands on a red number is$P\left(E\right)=\frac{9}{19}$.

(b)The probability that the wheel lands on a black number twice in a row is$P\left(E\right)=\frac{81}{361}$.

(c)The probability that the wheel lands on 0 or 00 is$P\left(E\right)=\frac{1}{19}$.

(d) The probability that in five spins the wheel never lands on either 0 or 00 is$P\left(E\right)=0.763122$.

(e) The probability that the wheel lands on one of the first six integers on one spin, but does not land on any of them on the next spin is$P\left(E\right)=\frac{48}{361}$ .

Step-by-step solution

The Concept of probability

If $\mathit{S}$represents the sample space and$\mathit{E}$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$.

The probability of winning a lottery (a)

Given that, in roulette, a wheel with 38 numbers is spun. Of these, 18 are red, and 18 are black. The other two numbers, which are neither black nor red, are 0 and 00.

The probability that when wheel is spun it lands on any particular number is$1/38$.

As per the problem we have been asked to find that the probability that the wheel lands on a red number.

If $S$represents the sample space and $E$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

As we know that total numbers spun is$38=n\left(s\right)$.

Event of wheel landing on a red number is$18=n\left(E\right)$.

Now substitute the values in the above formula we get,

$$\begin{gathered} P\left(E\right)=\frac{18}{38} \\ =\frac{9}{19} \end{gathered}$$

The probability that the wheel lands on a red number$P\left(E\right)=\frac{9}{19}$ .

The probability of winning a lottery (b)

Given that, in roulette, a wheel with 38 numbers is spun. Of these, 18 are red, and 18 are black. The other two numbers, which are neither black nor red, are 0 and 00.

The probability that when wheel is spun it lands on any particular number is$1/38$.

As per the problem we have been asked to find that the probability that the wheel lands on a black number twice in a row.

If $S$represents the sample space and $E$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

As we know that total numbers spun is$38=n\left(s\right)$.

Event of wheel landing on a black number is$18=n\left(E\right)$.

Now substitute the values in the above formula we get,

$$\begin{gathered} P\left(E\right)=\frac{18}{38} \\ =\frac{9}{19} \end{gathered}$$

Landing on a number second time is independent of the other events. So, the probability that the wheel lands on the black number twice in a row is,

$$\begin{gathered} P\left(E\right)=\frac{9}{19}\times \frac{9}{19} \\ =\frac{81}{361} \end{gathered}$$

The probability that the wheel lands on a black number twice in a row $P\left(E\right)=\frac{81}{361}$.

The probability of winning a lottery (c)

Given that, in roulette, a wheel with 38 numbers is spun. Of these, 18 are red, and 18 are black. The other two numbers, which are neither black nor red, are 0 and 00 .

The probability that when wheel is spun it lands on any particular number is$1/38$.

As per the problem we have been asked to find that probability that the wheel lands on 0 or 00.

If $S$represents the sample space and $E$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

As we know that total numbers spun is$38=n\left(s\right)$.

Event of wheel landing on 0 or 00 is$2=n\left(E\right)$.

Now substitute the values in the above formula we get,

$$\begin{gathered} P\left(E\right)=\frac{2}{38} \\ =\frac{1}{19} \end{gathered}$$

The probability that the wheel lands on 0 or 00 is $P\left(E\right)=\frac{1}{19}$.

The probability of winning a lottery (d)

Given that, in roulette, a wheel with 38 numbers is spun. Of these, 18 are red, and 18 are black. The other two numbers, which are neither black nor red, are 0 and 00.

The probability that when wheel is spun it lands on any particular number is$1/38$.

As per the problem we have been asked to find that probability that in five spins the wheel never lands on either 0 or 00.

If $S$represents the sample space and $E$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

As we know that total numbers spun is$38=n\left(s\right)$.

The wheel not landing on either 0or 00 means it can land on any other numbers. So, the event of wheel not landing on either 0 or 00 is,

$$\begin{gathered} n\left(E\right)=38-2 \\ =36 \end{gathered}$$.

Now substitute the values in the above formula we get,

$$\begin{gathered} P\left(E\right)=\frac{36}{38} \\ =\frac{18}{19} \end{gathered}$$

The probability that in one spin the wheel doesn't lands on either 0 or 00 is

$P\left(E\right)=\frac{18}{19}$.

The probability that in five spins the wheel never lands on either 0 or 00 is

role="math" localid="1668507834038" $$\begin{gathered} P\left(E\right)=\frac{18}{19}\times \frac{18}{19}\times \frac{18}{19}\times \frac{18}{19}\times \frac{18}{19} \\ P\left(E\right)=\frac{1889568}{2476099} \\ =0.763122 \end{gathered}$$

The probability that in five spins the wheel never lands on either 0 or 00 is

$P\left(E\right)=0.763122$.

The probability of winning a lottery (e) 

Given that, in roulette, a wheel with 38 numbers is spun. Of these, 18 are red, and 18 are black. The other two numbers, which are neither black nor red, are 0 and 00.

The probability that when the wheel is spun it lands on any particular number is$1/38$.

As per the problem we have been asked to find that the probability that the wheel lands on one of the first six integers on one spin, but does not land on any of them on the next spin.

If $S$represents the sample space and$E$ represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

As we know that total numbers spun is $38=n\left(s\right)$.

Number of ways the wheel lands on any of the number between 1 to 6 is$6=n\left(E_1\right)$ .

The probability that wheel lands on any of the number between 1 to 6 is

$$\begin{gathered} P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)} \\ =\frac{6}{38} \\ =\frac{3}{19}. \end{gathered}$$

Now, the number of ways the wheel lands on any number other than 1 to 6 is,

$$\begin{gathered} n\left(E_2\right)=38-6 \\ =32 \end{gathered}$$

The probability that wheel lands on any of the number other than 1 to 6 is

$$\begin{gathered} P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)} \\ =\frac{32}{38} \\ =\frac{16}{19} \end{gathered}$$

The probability that the wheel lands on one of the first six integers on one spin, but does not land on any of them on the next spin,

$$\begin{gathered} P\left(E\right)=P\left(E_1\right)\times P\left(E_2\right) \\ P\left(E\right)=\frac{3}{19}\times \frac{16}{19} \\ =\frac{48}{361} \end{gathered}$$

The probability that the wheel lands on one of the first six integers on one spin, but does not land on any of them on the next spin is$P\left(E\right)=\frac{48}{361}$