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(2) from the given data, The insurance company estimates that a total loss - RM Look will occur with probability 0.2% (0.002). we need to find the expected payment by insurance company is Expected loss = Sam (probability of loge loss value] 5900,000[1*0.662 +0.5 *0.01 +0.25 *oi] = 200.000 (0.032] = buso Ignoring all other partial loses the insurance company charge each year a profit of RM 500 of premium is premium amount = RM 6uro + RM Sto - RM 6900 Ang (3) let x be a poisson randon variable which denotes the number of accidents: x= 3 per month The probability that at most a accidents occur in a given = P(x=3) = P(x=0)+ P(x-1) +P(x=1) +P(X=3) - €3(1+3+ . for a given two months, average noot accidents = 3(2) = 6 To meet the gatety goal there ghould be no more than two accidents the probability of which is - 0.6472 -6 (1767 e 20.0620Any.
(4) probability of an electronic calculators to be faulty = 0.10 probability of an electronic calculator to be in working Condition = 0.90 when selected 5 calculators at Tandon, Cases that there will be more detective calculators compared to wooking ordered calculator will be Cage 1 3 detective and a working Cage 2 u detective and I working Cage 3: 5 defective and o working Before looking at the probabilities at uz tiset fanuliarulize augelves with the concept at binomial probability distribu- tien. In such cars we have a posebilities such that the Sun of probabilities of both the cases is let the probability of detective be p..
.
So out of n cases the detective probability that it will be for or no.ct cases will be ncr x prxli-pinet Now, let's look at their probabilities for this case! Cage 1 - probability = 5G X 0.10 0.90% = 0.0081 Cage 2: Papable+ : 5c, холочке-16 . o.eees Case 3: se probability = 5Cg *0:105x 0.18 = 0.0000! . .. Required probability = 0.0081 +0.oochs +0.0oool 0.00556 Ang Not: As por HomeworkLib rules, I have to do the angwer with in 2 hours and you posted the u questions. There is no time to do the ongewer from 1st question se, I don't have the time to draw the venn diagrams.
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