3.7.6 .WP A state runs a lottery in which six numbers are randomly selected from 40 without replacement. A player chooses six nu

Question

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3.7.6 .WP A state runs a lottery in which six numbers are randomly selected from 40 without replacement. A player chooses six nu

mbers before the state’s sample is selected. a. What is the probability that the six numbers chosen by a player match all six numbers in the state’s sample? b. What is the probability that five of the six numbers chosen by a player appear in the state’s sample? c. What is the probability that four of the six numbers chosen by a player appear in the state’s sample?

Mathematics

1 answer:

PIT_PIT [3.9K] · Answer 1 · 2026-02-25T10:07:35+03:00

Answer:

a) 2.60x10^-7

b) 5.31x10^-5

c) 2.19x10^-3

Step-by-step explanation:

X=the number of hits

Probability can be determined by taking the number of successful outcomes and dividing it by the total number of possible outcomes.

Now

a) P(X=6)=P({1, 1, 1, 1, 1, 1})=6/40*5/39*4/38*3/37*2/36*1/35=2.60x10^-7

b) P(X=5)=P({0, 1, 1, 1, 1, 1})+P({1, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 1, 0}), and each scenario carries the same probability

P(X=5)=6P({1, 1, 1, 1, 1, 0})=6*(6/40*5/39*4/38*3/37*2/36*34/35)=5.31x10^-5

c) P(X=4)=P({0, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 0, 0}), where each case has an identical probability.

$P(X=4)=(^6_4)P({1, 1, 1, 1, 0, 0})=\frac{6!}{4!(6-4)!}(\frac{6}{40} \frac{5}{39} \frac{4}{38} \frac{3}{37} \frac{34}{36} \frac{33}{35} )=2.19\times 10^{-3}$