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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.

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Answer:

[tex]\dfrac{1}{3,357,900}[/tex]

Step-by-step explanation:

There are

[tex]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[/tex]

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

[tex]\dfrac{1}{111,930}\cdot \dfrac{1}{30}=\dfrac{1}{3,357,900}[/tex]

p(win jackpot) =  [tex]\frac{1}{3357900}[/tex]

This question will be solved using the concept of probability, permutation and combination.

The number of ways to select four numbers between 1 and 42 will be by combination formula which is;

n[tex]C_{r}[/tex] = n!/(r!(n - r)!)

Thus, applying this to the question gives;

42[tex]C_{4}[/tex] = 42!/(4!(42 - 4)!)

This gives us 111930 ways

Now, we are told that you must also select the correct single number between 1 and 30. This means that the probability of winning the jackpot will be;

p(win jackpot) = [tex]\frac{1}{30}[/tex] × [tex]\frac{1}{111930}[/tex]

p(win jackpot) = [tex]\frac{1}{3357900}[/tex]

Read more at; brainly.com/question/17999498

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