A teacher gives her mathematics class $20$ study problems. She selects $10$ to answer on an upcoming test. A student can solve $15$ of the problems.
- Find the probability that the student can solve all $10$ problems on the test.
- Find the odds that the student will know how to solve $8$ of the problems.
$10$ problems can be selected from $20$ problems in
${^{20}}{C}_{10} = \dfrac{20!}{10! \times 10!} = 184756$ ways.
$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 184756$
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Let $A$ be the event that the student can solve all $10$ problems.
Total number of problems the student can solve $= 15$
$\therefore \;$ $10$ problems can be selected from $15$ problems in
${^{15}}{C}_{10} = \dfrac{15!}{10! \times 5!} = 3003$ ways
$\therefore \;$ Number of elements in $A = n \left(A\right) = 3003$
$\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{3003}{184756} = \dfrac{21}{1292}$
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Let $B$ be the event that the student solves $8$ problems.
Total number of problems the student cannot solve $= 5$
$8$ problems can be selected from $15$ problems in
${^{15}}{C}_{8} = \dfrac{15!}{8! \times 7!} = 6435$ ways
$2$ problems which cannot be solved can be selected from the $5$ problems in
${^{5}}{C}_{2} = \dfrac{5!}{3! \times 2!} = 10$ ways
$\therefore \;$ Number of ways in which the student can solve $8$ problems is
$= 6435 \times 10 = 64350$ ways
$\therefore \;$ Number of elements in $B = n \left(B\right) = 64350$
Let $P \left(s\right)$ be the probability of success of event $B$ and $P \left(f\right)$ be the probability of failure of event $B$
$P \left(s\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{64350}{184756} = \dfrac{225}{646}$
$P \left(f\right) = 1- \dfrac{225}{646} = \dfrac{421}{646}$
$\therefore \;$ Odds that the student will know how to solve $8$ of the problems
$= \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{225/646}{421/646} = \dfrac{225}{421}$