Question

In a certain state's lottery, 40 balls numbered 1 through 40 are placed in a machine and 7 of them are drawn at random. If the 7 numbers drawn match the 7 numbers a player has chosen, in any order, she or he win $2,500,000. Alternatively, if 6 of the numbers drawn match 6 of the 7 numbers a player has chosen, in any order, the player wins $40,000, and if 5 of the numbers drawn match 5 of the 7 numbers a player has chosen, in any order, the player wins $10,000.
Required:
a. What is the probability a player who buys one ticket will win the $2,500,000 prize?
b. What is the probability a player who buys one ticket will win the $10,000 prize?

Answer 1

Answer:

Step-by-step explanation:

From the given information;

The total number of ways to choose 7 number = $^{40}C_7$

Number of ways to choose 7 correct numbers = $^{7}C_7$

∴

The probability P( win $2500000) is;

$= \dfrac{^{7}C_7}{^{40}C_7}$

$= \dfrac{\dfrac{7!}{7!(7-7)!} }{\dfrac{40!}{7!(40-7)!}}$

$= \dfrac{1 }{\dfrac{40!}{7!(40-7)!}}$

$= \dfrac{1 }{\dfrac{40!}{7!(33)!}}$

$= \dfrac{1 }{18643560}$

= 5.36 × 10⁻⁸

The probability P( win $10000) is:

$= \dfrac{^7C_5 \times ^{33} C_2}{^{40}C_7}$

$= \dfrac{ \dfrac{7!}{5!(7-5)!} \times \dfrac{33!}{2!(33-2)!} }{ \dfrac{40!}{7!(40-7)!}}$

$= \dfrac{ \dfrac{7!}{5!(2)!} \times \dfrac{33!}{2!(31)!} }{ \dfrac{40!}{7!(33)!}}$

$= \dfrac{ 21 \times 528 }{ 18643560}$

$= \dfrac{ 11088 }{ 18643560}$

$=\dfrac{462}{776815}$

= 5.95 × 10⁻⁴