The provided solution to this question is: The probability needs to be calculate

Question

The provided solution to this question is:

The probability needs to be calculated for each total number of claims.

0: 0.5(0.2) = 0.10

1: 0.5(0.3) + 0.3(0.2) = 0.21

2: 0.5(0.4) + 0.3(0.3) + 0.2(0.2) = 0.33

3: 0.5(0.1) + 0.3(0.4) + 0.2(0.3) + 0.0(0.2) = 0.23

At this point there is only 0.13 probability remaining, so the mode must be at 2.

I'm not understanding how the fact that there's only 0.13 probability remaining, the mode is at 2? Please explain how we get that the mode is at 2 by the fact of 0.13 probability remaining.

6. An insurance company insures a good driver and a bad driver on the same policy. The table below gives the probability of a given number of claims occurring for each of these drivers in the next ten years Probability for the good driver 0.5 0.3 0.2 0.0 Number Probability for the bad driver 0.2 0.3 0.4 of claims The number of claims occurring for the two drivers are independent Calculate the mode of the distribution of the total number of claims occurring on this policy in the next ten years. (A) 0 (B)1 (C) 2 (D) 3 (E) 4

Explanation / Answer

here what above explanation trying to give is probability distribution of number of accidents.

P( 0 accidents) =P(0 accidents if good driver)*P(0 accidents if bad driver) =0.5*0.2 =1

P(1 accidents) =P(0 accidents if good driver)*P(1 accidents if bad driver) +P(1 accidents if good driver)*P(0 accidents if bad driver) = 0.5*0.3+0.3*0.2 =0.21

and so on.....

as we know that mode occurs at highest frequency or relative frequency :

for P(2 accidents) =0.33 ; and cause sum of probability is 1; tehrefore 0.13 probability remains to assign for 4,5,6 accidents

as 0.33 is higher than 0.13 ; tehrefore what they are trying to say that cause 0.13 can not excced 0.33 and 0.33 being maximum and corresponding to 2 accidents should be mode

(please revert for further clarification)

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