Question

The following table presents the probabilities for the number of times that a certain computer system will crash in the course of a week. Let A be the event that there are more than two crashes during the week, let B be the event that the system crashes at least once. Number of Crashes Probability 0.60 0.30 0.05 0.04 0.01 4 Find the sample space. Then find the subsets of the sample space that correspond to the events A and B. Then find P(A) and P(B).

Answer 1

Sample Space : All possible outcome of an event comes under sample space

For instance ;

in tossing 2 coins it's sample space = {(H,H) (H,T) (T,H) (T,T)}

Similarly ,in above case our sample space is {0,1, 2, 3, 4} because all these can occur as the computer may not be crash in week   or it may be crash 1 time in week    or 2 times or 3 times or 4 times in the week .

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Now, talking about subsets : A set whose each element must belongs to it's "parent set" or "super set"

So, the subset of the sample space correspond to event "A" : {3 , 4}

because the event "A" is for that computer must be crash more than 2 times during the week .

and the subset of the sample space corresponds to event "B" : {1 , 2 , 3 , 4}

because the event "B" is for that the computer must be crash at least once during the week .

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Now , let us discuss about Probability :

     $Probability = \frac{Total~Favorable~Outcome}{Total~Possible~Outcome}$

Therefore, for event "A" ,

Total Favorable outcome = $2$ (i.e. Number of crashes should be $3$ or $4$)

$\implies$   Total possible outcome = $5$(i.e. the computer may not be crash any time or it may crash for up to $4$ times)

Since, here their probabilities are already given we have to sum up all that .

$\implies$$P(A)= P(Crash~3~times) + P(Crash~4~times)$

$\implies$$P(A)= 0.04 + 0.01 = 0.05$ .

Similarly, For event "B" ,that computer must be crash at least once during the week that means it may be crash "1 time" or   "2 times" or "3 times" or "4 times"

Therefore, we will add up all given probabilities from crashing 1 times to 4 times

$\implies$$P(B) = P(Crash~1~time) + P(Crash~2~times) + P(Crash~3~times) + ~P(Crash~4~times)$

$\implies$  $P(B) = 0.30+0.05+0.04+0.01 = 0.4$ .