badgerboy
A certain telephone number has seven digits. If the telephone number has the digit zero exactly 3 times, and the number 1 is not used at all, what is the probability that the phone number contains one or more prime digits.
A. 1/24
B. 1/16
C. 1/2
D. 15/16
E. 23/24
telephone: {000}{XXXX}… X≠{0,1}={2,3,4,5,6,7,8,9}=8… primes={2,3,5,7}=4… not_prime,0,1={4,6,8,9}=4
METHOD 1: find the probability of not getting any prime and subtract from 1.
unfavorable outcomes {XXXX} not_prime,0,1: {4*4*4*4}=\(4^4\)
total outcomes {XXXX}: {8*8*8*8}=\(8^4\)
1-unfavorable/total: \(1-\frac{4^4}{8^4}=\frac{15}{16}\)
Answer (D)
METHOD 2: find the probability of each case with primes and add them.
favorable outcomes {PNNN}: {4*4*4*4}=\(4^4\) • arrangements {PNNN}: \(4!/3!=4\)… \(=4^5\)
favorable outcomes {PPNN}: {4*4*4*4}=\(4^4\) • arrangements {PPNN}: \(4!/2!2!=6\)…\(=4^4•6\)
favorable outcomes {PPPN}: {4*4*4*4}=\(4^4\) • arrangements {PPPN}: \(4!/3!=4\)… \(=4^5\)
favorable outcomes {PPPP}: {4*4*4*4}=\(4^4\) • arrangements {PPPP}: \(4!/4!=1\)…\(=4^4\)
total outcomes {XXXX}: {8*8*8*8}=\(8^4\)
sum probability each case: \(\frac{4^5+4^4•6+4^5+4^4}{8^4}=15/16\)
Answer (D)