A radio station runs a promotion at an auto show with a money box with 15 $50 ti

Question

A radio station runs a promotion at an auto show with a money box with 15 $50 tickets, 13 $25 tickets, and 15 $5 tickets. The box contains an additional 20 "dummy" tickets with no value. Three tickets are randomly drawn. Find the probability that all three tickets has no value.

A radio station runs a promotion at an auto show with a money box with 15 $50 tickets, 13 S25 tickets, and 15 $5 tickets. The box contains an additional 20 dummy tickets with no value. Three tickets ara randomly drawn. Find the probability that all three tickets have no value The probability that all three tickets drawn have no money value is Round to Sour decimal places as needed)

Explanation / Answer

there are total 15+13+15+20=63 tickets

we can draw 3 tickets from 63 tickets by 63C3 ways,

and we can draw  3 no value i.e. dummy tickets from 20 dummy tickets by 20C3 ways,

so here sample space size i.e. n(S)=63C3=39711

and favourable event i.e. n(E)=20C3=1140

required probablity=P(E)=n(E)/n(S)=20C3/63C3=1140/39711=0.0287

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