Could someone help me with these problems please? This is a closed-book, closed-

Question

Could someone help me with these problems please?

This is a closed-book, closed-notes exam, except for your notes on both side sof a 3'x5" card You may use a calculator. GOOD LUCK!! (Except as noted, each problem is worth 3 points) Find the indicated probability. 1) Some insect shell fragments were found in the ingredients for a production run of candy bars. Of 1) the bars produced in the run, 118 cans were found to be contaminated and 346 cans were not contaminated. If one of these bars is randomly selected, what is the probability of getting a contaminated bar? Round to the nearest thousandth, if necessary. A) 0.254 B) 0.341 D) 0.746 C) 0.008 7 k. sor 2) A researcher infects two lab rats with the ebola virus. Suppose the researcher treats both rats with 2) a new experimental drug, and observes whether each rat survives or dies. The simple events and probabilities of their occurrences are shown in the table (where S in the first position means that rat 1 survives, D in the first position means that rat 1 dies, etc.). Simple Events Probabilities 0.59 0.12 0.14 0.15 SD DS Find the probability that at least one of the rats does not survive. 0.41 A) 0.12 C) 0.15 D) 0.26

Explanation / Answer

1) The correct answer is A) 0.254

Solution

Let A denote the event that the inspected candy bar is contaminated with insect shell fragements.

Now, According to the question:

The total number of candy bars = 118 + 346 = 464

Now we have to find the probability that the inspected candy bar is contaminated with insect shell fragements i.e.

P(A) = Number of candy bars infected with with insect shell fragements / The total number of candy bars

=118 / 464

= 0.254 (Rounded off to nearest thousand)

Therefore the answer is A

2) The correct answer is B) 0.41

Solution

We have to find the probability

P(Atleast one of the rats did not survive)

= P(the first rat did not survive) + P(the second rat did not survive) + P(Both rats did not survive)

=P(DS) + P(SD) + P(DD)

= 0.14 + 0.12 + 0.15

= 0.41

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