asked 202k views
4 votes
A sample of 4 different calculators is randomly selected from a group

containing 46 that are defective and 26 that have no defects. What is the
probability that all four of the calculators selected are defective? Round to four
decimal places.
A) 0.1021 B) 0.1586 C) 0.1666 D) 10.9154

asked
User Lucasz
by
8.4k points

1 Answer

11 votes


\displaystyle\\|\Omega|=\binom{72}{4}=(72!)/(4!68!)=(69\cdot70\cdot71\cdot72)/(2\cdot3\cdot4)=1028790\\|A|=\binom{46}{4}=(46!)/(4!42!)=(43\cdot44\cdot45\cdot46)/(2\cdot3\cdot4)=163185\\\\P(A)=(163185)/(1028790)=(473)/(2982)\approx0.1586

answered
User Vcarel
by
9.2k points
Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.