Question

In a sample of 1000 adults, 150 said they are very confident in the nutritional information on restaurant menus. for us adults are selected at random without replacement. find the probability that all four us adults are very confident in the nutritional information on menus

Answer 1

There are 150 confident adults out of 1000.
Selecting four at random withOUT replacement
Probability that all four are confident
=(150/100*149/999*148/997*147/996)
=150!/146! / (1000!/996!)
= 7301/14925090
= 0.00048918

(check: answer < (150/1000)^4=0.00050625, ok)