A manufacturer of sports cars enters 3 drivers in a race Let

A manufacturer of sports cars enters 3 drivers in a race. Let A1 be the event that driver 1 \"shows\" (that is, he is among the first three drivers in the race to cross the finish line), let A2 be the event that driver 2 shows, and let A3 be the event that driver 3 shows. Assume that the events are independent and that .

P(A1) = P(A2) = P(A3) = 0.01

Compute the probability that (i) none of the driver will show, (ii) at least 1 will show, (iii) at least 2 will show, (iv) all of them will show.

Please include steps

Solution

Since the probability that a certain driver will show up is all the same, 0.01, then we can use the binomial distribution here.

i)

P(none) = (1- 0.01)^3 = 0.970299 [answer]

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ii)

Note that

P(at least 1) = 1 - P(0) = 1 - 0.970299

= 0.029701 [ANSWER]

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iii)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    3      
p = the probability of a success =    0.01      
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.999702
          
Thus, the probability of at least   2   successes is  
          
P(at least   2   ) =    0.000298 [ANSWER]

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IV)

P(all show) = 0.01^3 = 0.000001 [ANSWER]