An insurance company insure three type of vehicles ie., type A, B and

1.	An insurance company insure three type of vehicles ie., type A, B and C . If it insured 12000 vehicles of type A, 16000 vehicles of type B and 20,000 vehicles of type C. Survey report says that the chances of their accident are 0.01,0.03 and 0.04 respectively. Based on the informations given above, write the answer of following:(i) The probability of insured vehicle of type C(ii) Let E be the event that insured vehicle meets with an accident then P(E / A)(iii) Let E be the event that insured vehicle meets with an accident then P(E).(iv) The probability of an accident that one of the insured vehicle meets with an accident and it is a type C vehicle.
Answer» P(A) = \(\frac{n(A)}{n(S)}=\frac{12000}{48000}\) = \(\frac{12}{48}=\frac14\) P(B) = \(\frac{n(B)}{n(S)}=\frac{16000}{48000}\) = \(\frac{16}{48}=\frac13\) P(C) = \(\frac{n(C)}{n(S)}=\frac{20000}{48000}\) = \(\frac{20}{48}=\frac1{12}\) P(E/A) = 0.01, P(E/B) = 0.03 P(E/C) = 0.04 where E : The event of accident of vehicle. (i) P(C) = \(\frac{n(C)}{n(S)}=\frac{20000}{48000}=\frac{20}{48}=\frac5{12}\) (ii) P(E/A) = 0.01 (iii) P(E) = P(E/A)P(A) + P(E/B)P(B) + P(E/C) P(C) = 0.01 x 1/4 + 0.03 x 1/3 + 0.04 x 5/12 = 1/400 + 1/100 + 5/300 = \(\frac{3+12+20}{1200}\) = \(\frac{35}{1200}\) = \(\frac{7}{240}\) (iv) P(C/E) = \(\frac{P(E/C)P(C)}{P(E/A)P(A)+P(E/B)P(B)+P(E/C)P(C)}\) = \(\cfrac{0.04\times\frac5{12}}{\frac{35}{1200}}=\cfrac{\frac{20}{1200}}{\frac{35}{1200}}\) = \(\frac{20}{35}=\frac47\)

An insurance company insure three type of vehicles ie., type A, B and C . If it insured 12000 vehicles of type A, 16000 vehicles of type B and 20,000 vehicles of type C. Survey report says that the chances of their accident are 0.01,0.03 and 0.04 respectively. Based on the informations given above, write the answer of following:(i) The probability of insured vehicle of type C(ii) Let E be the event that insured vehicle meets with an accident then P(E / A)(iii) Let E be the event that insured vehicle meets with an accident then P(E).(iv) The probability of an accident that one of the insured vehicle meets with an accident and it is a type C vehicle.

Answer»

P(A) = \(\frac{n(A)}{n(S)}=\frac{12000}{48000}\) = \(\frac{12}{48}=\frac14\)

P(B) = \(\frac{n(B)}{n(S)}=\frac{16000}{48000}\) = \(\frac{16}{48}=\frac13\)

P(C) = \(\frac{n(C)}{n(S)}=\frac{20000}{48000}\) = \(\frac{20}{48}=\frac1{12}\)

P(E/A) = 0.01, P(E/B) = 0.03

P(E/C) = 0.04

where E : The event of accident of vehicle.

(i) P(C) = \(\frac{n(C)}{n(S)}=\frac{20000}{48000}=\frac{20}{48}=\frac5{12}\)

(ii) P(E/A) = 0.01

(iii) P(E) = P(E/A)P(A) + P(E/B)P(B) + P(E/C) P(C)

= 0.01 x 1/4 + 0.03 x 1/3 + 0.04 x 5/12

= 1/400 + 1/100 + 5/300 = \(\frac{3+12+20}{1200}\) = \(\frac{35}{1200}\) = \(\frac{7}{240}\)

(iv)
P(C/E) = \(\frac{P(E/C)P(C)}{P(E/A)P(A)+P(E/B)P(B)+P(E/C)P(C)}\)

= \(\cfrac{0.04\times\frac5{12}}{\frac{35}{1200}}=\cfrac{\frac{20}{1200}}{\frac{35}{1200}}\) = \(\frac{20}{35}=\frac47\)