Question #105393
3) Our class is creating a new academic team, the Data Managers. The team will have 4 members to perform different tasks: collector, analyzer, grapher, and recorder. 10 people in class want to be on the team, so we pick a random team out of the 10.
A) If exactly 4 boys volunteer, what is the probability that the team is all boys or no boys? Evaluate your answer.
B) What is the probability that Student 1 and Student 2 are on the team? Evaluate your answer.
C) With two teacher coaches, the team of four takes a picture (6 people in picture). What is the probability that the teachers are together in the picture? Evaluate your answer.
1
Expert's answer
2020-03-16T10:46:26-0400

A) If exactly 4 boys volunteer, what is the probability that the team is all boys or no boys? 


P(all four boys or no boys)=P(all\ four\ boys\ or\ no\ boys)=

=P(all four boys)+P(no boys)==P(all\ four\ boys)+P(no\ boys)=

=(44)(10444)(104)+(40)(10440)(104)=={\dbinom{4}{4}\dbinom{10-4}{4-4} \over \dbinom{10}{4}}+{\dbinom{4}{0}\dbinom{10-4}{4-0} \over \dbinom{10}{4}}=

=1(1)10!4!(104)!+16!4!(64)!10!4!(104)!=={1(1) \over \dfrac{10!}{4!(10-4)!}}+{1\cdot\dfrac{6!}{4!(6-4)!} \over \dfrac{10!}{4!(10-4)!}}==1210+15210=8105={1 \over 210}+{15 \over 210}={8 \over 105}


B) What is the probability that Student 1 and Student 2 are on the team?


P(Student1 & Student2)=(11)(10242)(104)=P(Student1\ \&\ Student2 )={\dbinom{1}{1}\dbinom{10-2}{4-2} \over \dbinom{10}{4}}==18!2!(82)!10!4!(104)!=28210=215={1\cdot\dfrac{8!}{2!(8-2)!} \over \dfrac{10!}{4!(10-4)!}}={28 \over 210}={2 \over 15}



C) With two teacher coaches, the team of four takes a picture (6 people in picture). What is the probability that the teachers are together in the picture? 

Consider two teacher coaches as one unit.


P(teachers are together)=2!5!6!=26=13P(teachers\ are \ together)={2!\cdot5! \over 6!}={2 \over 6}={1 \over 3}




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