Question

Commercial aircraft used for flying in instrument conditions are required to have two independent radios instead of one. Assume that for a typical flight, the probability of a radio failure is 0.0023. What is the probability that a particular flight will be safe with at least one working radio? Why does the usual rounding rule of three significant digits not work here? Is this probability high enough to ensure flight safety? The probability that at least one of the radios works correctly is (Round to six decimal places as needed.)

Answer 1

Answer)

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 1 - 0.0023 = 0.9977

N = number of trials = 2

R = desired success = at least 1.

P(at least 1) = 1 - P(0) = 0.99999471

C. Rounding the result to three significant digits yields 1.00, which is misleading because it suggests that it is certain that both radios will work correctly.

Part 2)

From z table, P(z>1.34) = 0.0901

P-value = 0.0901

From z table, P(z>1.645) = 0.05

So, critical value is 1.645