Thursday Friday and Saturday are the busiest nights at the J

Thursday, Friday, and Saturday are the busiest nights at the Jopper bar. Police records show that on 12 of the last 50Thursdays, 15 of the last 50 Fridays, and 16 of the last Saturdays they were summoned to deal with a disturbance at the bar. Construct a tree diagram and use it to find the probability that over the next Thursday, Friday, and Saturday nights there will be:

a) no trouble

b) trouble on Thursday only

c) trouble on one night only

d) trouble on Friday and Saturday only

e) trouble on two nights only

Solution

Let Th shows the event that there is a trouble on Thursday, F shows the event that there is a trouble on Friday, and S shows the event that there is trouble on Saturday. So we have follwoing probabilties:

P(Th)=12/50=0.24

P(F)=15/50=0.3

P(S)=16/50=0.32

(a)

Since trouble on one day is independent from the trouble of other day so the probability that over the next Thursday, Friday, and Saturday nights there will be no trouble is

[1-P(Th)]*[1-P(F)]*[1-P(S)]=[1-0.24]*[1-0.3]*[1-0.32]=0.36176

Hence, the required probability is 0.3618.

(b)

That is there will no trouble on Friday and Saturday. So the required probability is

[P(Th)]*[1-P(F)]*[1-P(S)]=[0.24]*[1-0.3]*[1-0.32]=0.11424

hence, the required probability is 0.1142.

(c)

That is there will be trouble either on Thrusday or Friday or Saturday.The probability that there will be trouble on Thrusday only is

[P(Th)]*[1-P(F)]*[1-P(S)]=[0.24]*[1-0.3]*[1-0.32]=0.11424

The probability that there will be trouble on Friday only is

[1-P(Th)]*[P(F)]*[1-P(S)]=[1-0.24]*[0.3]*[1-0.32]=0.15504

The probability that there will be trouble on Saturday only is

[1-P(Th)]*[1-P(F)]*[P(S)]=[1-0.24]*[1-0.3]*[0.32]=0.17024

Hence, the required probability is

0.11424+0.15504+0.17024=0.43952

(d)

That is there is no trouble on Thursday so

[1-P(Th)]*[P(F)]*[P(S)]=[1-0.24]*[0.3]*[0.32]=0.07296

Hence, the required probability is 0.07296.

(e)

The probbaility that there is trouble on Friday and Saturday only is

[1-P(Th)]*[P(F)]*[P(S)]=[1-0.24]*[0.3]*[0.32]=0.07296

The probbaility that there is trouble on Thursday and Friday only is

[P(Th)]*[P(F)]*[1-P(S)]=[0.24]*[0.3]*[1-0.32]=0.04896

The probbaility that there is trouble on Thursday and Saturday only is

[P(Th)]*[1-P(F)]*[P(S)]=[0.24]*[1-0.3]*[0.32]=0.05376

Hence, the required probability is

0.07296+0.04896+0.05376=0.17568