document.write( "Question 1163248: The probability that a candidate passes a certain professional examination is
\n" ); document.write( "0.7. Find the probability that a candidate would pass the examination before
\n" ); document.write( "the fourth attempt. \n" ); document.write( "

Algebra.Com's Answer #787272 by Theo(13342) $\"\"$ $\"About$

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if the probability to pass the exam is .7, then the probability to fail the exam is 1 - .7 = .3
\n" ); document.write( "in order to not pass the exam before the fourth attempt, he would have to fail the exam the first 3 times he took it.
\n" ); document.write( "the probability of that happening would be .3 * .3 * .3 = .027, therefore the probability that it doesn't happen is 1 - .027 = .973.
\n" ); document.write( "that's the probability that he would pass the exam before the fourth attempt.
\n" ); document.write( "you can also look at it another way.
\n" ); document.write( "the probability that he would pass the exam before the fourth attempt would be equal to:
\n" ); document.write( ".7 if he passed it on the first attempt.
\n" ); document.write( ".3 * .7 if he passed it on the second attempt.
\n" ); document.write( ".3 * .3 * .7 if he passed it on the third attempt.
\n" ); document.write( "add those up and you get .7 + .21 + .063 = .973
\n" ); document.write( "the numbers check out.
\n" ); document.write( "your solution appears to be:
\n" ); document.write( "the probability that he would pass the exam before the fourth attempt is .973.\r
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