Question

Winning the jackpot in a particular lottery requires that you select the correct four numbers between 1 and 42 ​and, in a separate​ drawing, you must also select the correct single number between 1 and 30. Find the probability of winning the jackpot.

Answer 1

Answer:

Probability of winning the jackpot = 0.000000297

Step-by-step explanation:

Here, there are two sequence of events

First event (m) - choosing four  number between $1$ and $42$

Second event (n) - Select a single number between $1$ and $30$

Thus, the number of ways the two events can occur is

$m * n$

Number of ways the first event can occur is equal to

$^{42}C_4$

On solving we get

$\frac{42!}{4! * 38!}\\ = \frac{42*41*40*39*38!}{4! * 38!} \\= \frac{42*41*40*39}{4*3*2*1}$

$= 111930$ ways

The number of ways the two events can occur is

$111930 * 30 = 3357900$ ways

Probability of winning the jackpot

$\frac{1}{3357900}$

Probability of winning the jackpot = 0.000000297

Answer 2

Answer:

$\dfrac{1}{3,357,900}$

Step-by-step explanation:

There are

$C^{42}_4=\dfrac{42!}{4!(42-4)!}=\dfrac{42!}{4!\cdot 38!}=\dfrac{38!\cdot 39\cdot 40\cdot 41\cdot 42}{2\cdot 3\cdot 4\cdot 38!}=39\cdot 10\cdot 41\cdot 7=111,930$

different ways to select the  four numbers between 1 and 42. Only one of this ways is correct (successful to win).

There are 30 different ways select the single number between 1 and 30. Only one of them is correct.

The  probability of winning the jackpot is

$\dfrac{1}{111,930}\cdot \dfrac{1}{30}=\dfrac{1}{3,357,900}$