Question

Surveys have indicated that in a given region 79% of car occupants wear seat belts regardless of where they sit in the car. Wearing of seat belts in the region is random and shows no regular pattern. The surveys have shown also that in 45% of cars the driver is the sole occupant, in 25% there are two occupants, in 15% three occupants, in 10% four occupants and in 5% five occupants. a) What is the probability that a car picked at random will have exactly three persons not wearing their seat belts? b) What is the probability that of three cars chosen at random, exactly two have all occupants wearing seat belts?

Answer 1

sol lets draw in problem a free diagram and solve the representing the information given questions using free diagram. 0.45 0.25 / 0.15 0.0 Sole occupant Two occupants Three r five оссарель four оссор, осачрель 0.797 10.79 0.79) 1-0.79 0.79 10.21 0.79) 0.79 10.2 0.7 10.21 -0.21 0.797 10.21 0.79 I do not wear wear klear seatbelt I do not seak Weay ܐܘܢ lear rdo not wear) Seat I wear Seat belt I seat beit belt! ar reat seat belt Seat rdo not weay rcat belt belt belt beit a) The probability that a car picked at random will have exactly 3 persons not wearing seat bels. Here, we should not consider first two stems of free diagram [sole occupants, Two occupants as exactly Free persons not So, we have to consider 3 ocupants, four occupant and five occupants stems .

Required probability: (from Freediagram] (0.15) all Ihree of three occupants not wearing seat belts] + (6. 10) (three of four occupants not wearing seat belts] + (o.os) [base of tive occupants not wearing sealt belts] - (0.15) (0.21] + (0-10) 45 (0.213(0.79)] +0.05 [556.25% (0.79 - (0.15) (0.009201) + (0-10) (0.02926476) +0.05) (0.057797901) 0.00 138915 o.o02926476+ 0.00288 98 9505 = 0.00720552105 x 0.0072 the probability that in first we need to find a car Here, choren at random, all occupants are wearing near from free diagran, this probability in (0.45) (0.79) + (0.25) (0.79) + (0.15) 10.79) + (0-1) (0.79) + (0.05) 6-79) - 0.63981621299 Now let x= No. of ocars with all occupants wearing seat belts out of 3 cars chosen at random n=3, p= 0.63981 (21299 X~ Binomial (n=3, p = 0.63981621299) Here, we need to find P(X=2) P(X=2) = 137 (0.63981621299) (aces 1-0.63981 621299) = 0.44233967711