Question:

(a) To determine the probability that the player wins the jackpot.

(b)To determine the probability that the player wins 1000000$, the prize for matching the first five numbers, but not the sixth number drawn.

(c)To determine the probability that a player win 500$, the prize for matching exactly four of the first five numbers, but not the sixth number drawn.

(d) To determine the probability that a player wins 10$, the prize for matching exactly three of the first five numbers but not the sixth number drawn, or for matching exactly two of the first five numbers and the sixth number drawn.

(a) Given that, a player in the Mega lottery picks five different integers between 1 and 70. inclusive, and a sixth integer between 1 and 25, inclusive, which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn.

(b) Given that, a player in the Mega lottery picks five different integers between 1 and 70. inclusive, and a sixth integer between 1 and 25, inclusive, which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn.

(c) Given that, a player in the Mega lottery picks five different integers between 1 and 70. inclusive, and a sixth integer between 1 and 25, inclusive, which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn.

(d) Given that, a player in the Mega lottery picks five different integers between 1 and 70. inclusive, and a sixth integer between 1 and 25, inclusive, which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn.

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

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Given that, a player in the Mega lottery picks five different integers between 1 and 70. inclusive, and a sixth integer between 1 and 25, inclusive, which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn.

As per the problem we have been asked to find that the probability that the player wins the jackpot.

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:

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

Number of ways of picking 5 integers from 70 integers is $70C_5$

And Number of ways of picking 1 integer from 25 integers is $25C_1$.

Sample space, $n\left(S\right)=70C_5\times 25C_1$

The player wins the jackpot if all six numbers match the numbers drawn.

Event of winning the jackpot is$n\left(E\right)=6C_6$ .

Now substitute the values in the above formula we get,

$P\left(E\right)=\frac{6C_6}{70C_5\times 25C_1}=\frac{1}{302575350}$

The probability that the player wins the jackpot is $P\left(E\right)=\frac{1}{302575350}$.

Given that, a player in the Mega lottery picks five different integers between 1 and 70. inclusive, and a sixth integer between 1 and 25, inclusive, which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn.

As per the problem we have been asked to find that the probability that the player wins 1000000$, the prize for matching the first five numbers, but not the sixth number drawn.

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:

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

Number of ways of picking 5 integers from 70 integers is $70C_5$.

Event of matching all five numbers drawn $=5C_5=1$

Probability that the first five numbers match, $P\left(E_1\right)=\frac{5C_5}{70C_5}$

And Number of ways of picking sixth integer from 25 integers is $25C_1$.

Event that the sixth integer matches $1C_1$

Probability that the sixth integer matches, $\frac{1C_1}{25C_1}$

Probability that the sixth integer doesn't matches, role="math" localid="1668512246968" $P\left(E_2\right)=1-\frac{1C_1}{25C_1}$

Required Probability that player wins 1000000$,

$$\begin{gathered} P\left(E\right)=P\left(E_1\right)\times P\left(E_2\right) \\ P\left(E\right)=\left(\frac{5C_5}{70C_5}\right)\times \left(1-\frac{1C_1}{25C_1}\right) \\ P\left(E\right)=\frac{1}{12607306} \end{gathered}$$

The probability that the player wins 1000000$ the prize for matching the first five numbers, but not the sixth number drawn, $P\left(E\right)=\frac{1}{12607306}$

Given that, a player in the Mega lottery picks five different integers between 1 and 70. inclusive, and a sixth integer between 1 and 25, inclusive, which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn.

As per the problem we have been asked to find that the probability that a player win 500$, the prize for matching exactly four of the first five numbers, but not the sixth number drawn.

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:

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

Number of ways of picking 5 integers from 70 integers is$70C_5$ .

Event of matching exactly four of the first five numbers drawn $=5C_4$

Event of matching remaining one number from rest 65 integers $=65C_1$

Probability that the first four of the five numbers drawn matches, $P\left(E_1\right)=\frac{5C_4\times 65C_1}{70C_5}$

And Number of ways of picking sixth integer from 25 integers is$25C_1$ .

Event that the sixth integer matches $1C_1$

Probability that the sixth integer matches,$\frac{1C_1}{25C_1}$

Probability that the sixth integer doesn't matches, data-custom-editor="chemistry" $P\left(E_2\right)=1-\frac{1C_1}{25C_1}$

Required Probability that player wins 500$,

$$\begin{gathered} P\left(E\right)=P\left(E_1\right)\times P\left(E_2\right) \\ P\left(E\right)=\left(\frac{5C_4\times 65C_1}{70C_5}\right)\times \left(1-\frac{1C_1}{25C_1}\right) \\ P\left(E\right)=\frac{52}{2017169} \end{gathered}$$

The probability that a player win 500$, the prize for matching exactly four of the first five numbers, but not the sixth number drawn, $P\left(E\right)=\frac{52}{2017169}$.

Given that, a player in the Mega lottery picks five different integers between 1 and 70. inclusive, and a sixth integer between 1 and 25, inclusive, which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn.

As per the problem we have been asked to find that the probability a player wins $\$ 10$, the prize for matching exactly three of the first five numbers but not the sixth number drawn, or for matching exactly two of the first five numbers and the sixth number drawn.

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:

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

Number of ways of picking 5 integers from 70 integers is $70C_5$.

Event of matching exactly three of the first five numbers drawn $=5C_3$

Event of matching remaining two number from rest 65 integers $=65C_2$

Probability that the first four of the five numbers drawn matches, $P\left(E_1\right)=\frac{5C_3\times 65C_2}{70C_5}$

And Number of ways of picking sixth integer from 25 integers is$25C_1$ .

Event that the sixth integer matches $1C_1$

Probability that the sixth integer matches,$\frac{1C_1}{25C_1}$

Probability that the sixth integer doesn't matches, $P\left(E_2\right)=1-\frac{1C_1}{25C_1}$

Probability that the exactly three of the first five numbers but not the sixth number drawn matches,

role="math" localid="1668513008513" $$\begin{gathered} P\left(E_a\right)=P\left(E_1\right)\times P\left(E_2\right) \\ P\left(E_a\right)=\left(\frac{5C_3\times 65C_2}{70C_5}\right)\times \left(1-\frac{1C_1}{25C_1}\right) \\ P\left(E_a\right)=\frac{1}{606} \end{gathered}$$

Similarly, the probability for matching exactly two of the first five numbers and the sixth number drawn,

$$\begin{gathered} P\left(E_b\right)=\left(\frac{5C_2\times 65C_3}{70C_5}\right)\times \left(\frac{1C_1}{25C_1}\right) \\ P\left(E_b\right)=\frac{1}{693} \end{gathered}$$

So, required probability that a player wins 10$,

$$\begin{gathered} P\left(E\right)=P\left(E_a\right)+P\left(E_b\right)\backslash \\ P\left(E\right)=\left(\frac{1}{606}\right)+\left(\frac{1}{693}\right) \\ P\left(E\right)=\frac{433}{139986} \end{gathered}$$

The probability that a player wins 10$, the prize for matching exactly three of the first five numbers but not the sixth number drawn, or for matching exactly two of the first five numbers and the sixth number drawn, $P\left(E\right)=\frac{433}{139986}$.