Answer :
Answer:
The probability that Calvin will get a ticket on any one morning is 0.27.
Step-by-step explanation:
Denote the events as follows:
X = Calvin will get a ticket
B = trap on route B is operated when a person is speeding
C = trap on route C is operated when a person is speeding
D = trap on route D is operated when a person is speeding
The information provided is:
[tex]P (X|A)=0.40\\P(X|B)=0.30\\P(X|C)=0.20\\P(X|D)=0.30\\P(A)=0.20\\P(B)=0.10\\P(C)=0.50\\P(D)=0.20[/tex]
The law of total probability states that:
[tex]P(A)=P(A|X_{1})P(X_{1})+P(A|X_{2})P(X_{2})+...+P(A|X_{n})P(X_{n})[/tex]
Use this law to compute the probability that Calvin will get a ticket on any one morning as follows:
[tex]P(X)=P(X|A)P(A)+P(X|B)P(B)+P(X|C)P(C)+P(X|D)P(D)\\=(0.40\times0.20)+(0.30\times0.10)+(0.20\times 0.50)+(0.30\times 0.20)\\=0.27[/tex]
Thus, the probability that Calvin will get a ticket on any one morning is 0.27.
Answer:
Probability that Calvin will get a ticket on any one morning is 0.27.
Step-by-step explanation:
We are given that the traps on routes A, B, C, and D are operated 40% , 30%, 20%, and 30% of the time, respectively. Calvin always speeds to class, and he has probability 0.2, 0.1, 0.5, and 0.2 of using those routes.
Firstly, Let Probability that traps are operated on route A = P(A) = 0.40
Probability that traps are operated on route B = P(B) = 0.30
Probability that traps are operated on route C = P(C) = 0.20
Probability that traps are operated on route D = P(D) = 0.30
Now, Let S = event of Calvin speeding to class
So, Probability that Calvin speeds to class given that he used route A = P(S/A) = 0.2
Probability that Calvin speeds to class given that he used route B = P(S/B) = 0.1
Probability that Calvin speeds to class given that he used route C = P(S/C) = 0.5
Probability that Calvin speeds to class given that he used route D = P(S/D) = 0.2
Now, the law of total probability states that ;
[tex]P(Z)=P(A|Z_{1})P(Z_{1})+P(A|Z_{2})P(Z_{2})+...+P(A|Z_{n})P(Z_{n})[/tex]
So, Probability that he will get a ticket on any one morning is given by;
= P(A) P(S/A) + P(B) P(S/B) + P(C) P(S/C) + P(D) P(S/D)
= [tex]0.40 \times 0.2 + 0.30 \times 0.1 + 0.20 \times 0.5 + 0.30 \times 0.2[/tex]
= 0.08 + 0.03 + 0.1 + 0.06 = 0.27
Therefore, Probability that he will get a ticket on any one morning is 0.27.