document.write( "Question 1049372: A pro basketball player is a poor​ free-throw shooter. Consider situations in which he shoots a pair of free throws. The probability that he makes the first free throw is 0.49. Given that he makes the​ first, suppose the probability that he makes the second is 0.65. Given that he misses the​ first, suppose the probability that he makes the second one is 0.41. Complete parts a through c below.\r
\n" ); document.write( "\n" ); document.write( "possiblity of him making both free throws is .49*.65 .3185 \r
\n" ); document.write( "\n" ); document.write( "B. Find the probability that he makes one of the two free throws using the multiplicative rule with the two possible ways he can do this.
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Algebra.Com's Answer #664901 by Aevans52(1)\"\" \"About 
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P(3 Free Throws Made) = P(One Free Throw Made AND One Free Throw Made AND One Free Throw Made) \r
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\n" ); document.write( "\n" ); document.write( "P(3 Free Throws Made) = P(One Free Throw Made) * P(One Free Throw Made) * P(One Free Throw Made) \r
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\n" ); document.write( "\n" ); document.write( "P(3 Free Throws Made) = (7/10) * (7/10) * (7/10) \r
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\n" ); document.write( "\n" ); document.write( "P(3 Free Throws Made) = (7*7*7)/(10*10*10) \r
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\n" ); document.write( "\n" ); document.write( "P(3 Free Throws Made) = 343/1000 \r
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\n" ); document.write( "\n" ); document.write( "P(3 Free Throws Made) = 0.343 \r
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\n" ); document.write( "\n" ); document.write( "So the probability is 343/1000 or 0.343 (which is a 34.3% chance) to make all 3 free throws. \r
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