I give quick best answers thank you for your help 1 Four st

I give quick best answers, thank you for your help! :)

1. Four students, Rick, Robin, Ryu and Roy are registered for the same class and attend independently of one another, Rick 98.1% of the time, Robin 91.4% of the time, Ryu 88.4% of the time, and Roy 82.5% of the time. What is the probability that on any given day:

a) That at least three of them will be in class?

b) That none of them will be in class?

Solution

Probability of rick, P(RK)=0.981,

Probability of not rick =1-0.981=0.019

Probability of Robin, P(RN)=0.914

Probability of not Robin=1-0.914=0.086

Probability of Ryu , P(RU)=0.884

Probability of not Ryu =1-0.884=0.116

Probability of Roy , P(RY)=0.825

Probability of not Roy=1-0.825=0.175

a) Probability that on any given day at least three of them will be in class

=Probability all four in class+Probability of Rick-Robin-RYu in class+Probability of Robin, Ryu, Roy, in class+Probability of Ryu, Roy, RIck, in class+Probability of Rick, Robin, Roy, in class

=0.981*0.914*0.884*0.825+0.981*0.914*0.884+0.914*0.884*0.825+0.884*0.825*0.981+0.825*0.981*0.914

=0.6539+0.7926+0.6665+0.7514+0.7397

=3.6041

b) That none of them will be in class=0.019*0.086*0.116*0.175=3.31702*10^-5