Choice leads for developing new business are randomly assigned to 58 employees w
ID: 3352057 • Letter: C
Question
Choice leads for developing new business are randomly assigned to 58 employees who make up the direct sales team. Half of the sales team is male, and half is female. An employee can receive at most one choice lead per day. On a particular day, five choice leads are assigned.
a. Are the events [first lead is to a female] and [second lead is to a female ] dependent or independent?
b. If the first four leads all go to women, what is the probability that the fifth lead also goes to a woman?
c. What is the probability that all five leads go to women if you know that at least four of the leads go to women?
Explanation / Answer
a) Event A: first lead is to a female
Event B: second lead is to a male
number of males =58/2 =29
number of females =29
total =29+29=58
P(A) = 29/58 =0.5
P(B) =29/58 =0.5
P(AnB) =(29/58)*(29/57) =841/3306 =0.254
For independent events,P(AnB)=P(A)*P(B)
0.254=0.5*0.5
0.254 =0.25
CLearly,P(AnB) is not equal to P(A)*p(B)
Hence the events are dependent events
b) since first four leads go to women,we are left with number of women =29-4 =25
and total people left =58-4 =54
so probability that fifth lead is women = 25/54 =0.463
c)
favourable ways = all leads go to female = 29C5 =118755
total ways = atleast 4 goes to female = 4 goes to female and 1 goes to male + all 5 goes to female
= 29C4*29C1 +29C5
=23751*29 +118755
=688779+118755
=807534
So prob =favourable ways/total ways
=118755/807534
=5/34
=0.147
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