document.write( "Question 938539: Commercial aircraft used for flying in instrument conditions must have two independent radios instead of one. Assume that for a typical flight, the probability of a radio failure is 0.0047. 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?
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Algebra.Com's Answer #571787 by mathmate(429) $\"\"$ $\"About$

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Failure rate for each instrument, p = 0.0047
\n" ); document.write( "It is a requirement that the two instruments are independent, which means that the probability that both will fail is governed by the multiplication rule, i.e.
\n" ); document.write( "Probability of both (independent) instruments fail
\n" ); document.write( "= p*p = $\"0.0047%5E2\"$
\n" ); document.write( "= $\"2.209+10%5E%28-5%29\"$
\n" ); document.write( "Probability that at least one instrument will be functional
\n" ); document.write( "= 1-p*p
\n" ); document.write( "= 0.99997791
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\n" ); document.write( "Rounding to three significant figures will be misleading because the correct answer is very close to 1. By rounding to three significant figures the result is 1.000, which has a very different interpretation from that of the correct answer.
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\n" ); document.write( "I do not know the required safety standard. Generally if p is small enough, doubling with an independent instrument will reduce the probability of both instruments failing to very small value. \n" ); document.write( "

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