Question

In a lottery game, a player picks six numbers from 1 to 48. If 4 of those 6 numbers match those drawn, the player wins third prize. What is the probability of winning this prize?

Answer 1

$|\Omega|={_{48}C_6}=\dfrac{48!}{6!42!}=\dfrac{43\cdot44\cdot45\cdot46\cdot47\cdot48}{720}=12271512\\|A|={_6C_4}=\dfrac{6!}{4!2!}=\dfrac{5\cdot6}{2}=15\\\\P(A)=\dfrac{15}{12271512}=\dfrac{5}{4090504}\approx1.2\%$

Answer 2

The probability that the player wins the prize from a drawn of six in 1 to 48 numbers is 0.001052.

Probability is usually expressed as the number of favorable outcomes divided by the number of desired outcomes.

Mathematically;

Probability of winning = $\mathbf{\dfrac{favorable \ outcomes}{number \ of \ desired \ outcomes}}$

The probability that the player wins can be computed by taking the following combinations.

From the given information;

• the number of ways in which 4 winning numbers can be selected out of six = $\mathbf{^6C_4}$

• the number of ways in which 2 non-winning numbers can be selected out of the 42 non-winning numbers i.e (48 - 6 = 42 ) = $\mathbf{^{42}C_2}$

• the number of ways to pick 6 numbers out of 48 = $\mathbf{^{48}C_6}$

Thus, the probability of winning can now be computed as:

$\mathbf{P(winning \ prize) = \dfrac{^6C_4 \times ^{42}C_2}{^{48}C_6}}$

$\mathbf{P(winning \ prize) = \dfrac{\Big (\dfrac{6!}{4!(6-4)!} \Big )\times \Big (\dfrac{42!}{2!(42-2)!} \Big )}{\Big (\dfrac{48!}{6!(48-6)!} \Big )}}$

$\mathbf{P(winning \ prize) = \dfrac{12915}{12271512}}$

$\mathbf{P(winning \ prize) =0.001052}$

Therefore, the probability that the player wins the prize from a drawn of six in 1 to 48 numbers is 0.001052.

Learn more about probability here:

https://brainly.com/question/11234923?referrer=searchResults