Question

A contest consists of predicting the results (win, draw or defeat) of 10 football matches. The probability that one entry contains at least 5 correct answers is $$$$
A. \dfrac{12585}{{{3}^{10}}}$$$$$ B. \dfrac{12385}{{{3}^{10}}} C. $\dfrac{9335}{{{3}^{10}}}
D. $\dfrac{12496}{{{3}^{10}}}$$$$$

Answer 1

We first find the number of all possible outcomes $n\left( S \right)$ by counting the number of ways we can fill an entry of 10 matches with three outcomes win, lose or draw. We find the number of favourable outcomes $n\left( A \right)$ by counting the number of ways we can select at least 5 that means 5,6,7,8,9,10 matches out of 10 and then filing the entry with one correct and two possible incorrect guesses. We find the required probability $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$$$$$

Complete step-by-step answer:
We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring or number of favourable outcomes and $n\left( S \right)$ is the size of the sample space or number of all possible outcomes then the probability of the event $A$ occurring is given by
$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$
We are given the question that the contest consists of predicting the results (win, draw or defeat) of 10 football matches. We see there are 3 outcomes for each match win, draw or loss. So the sample size is the total number of outcomes that is
$n\left( S \right)=3\times 3\times ...\left( 10\text{ times} \right)={{3}^{10}}$
Now we need to find the number of favourable outcomes of the event $A$ of guessing at least 5 correct entries. We see that at least 5 matches out of 10 matches have the correct answer in the entry. So we can choose matches with correct entries in one of ${}^{10}{{C}_{5}},{}^{10}{{C}_{6}},...,{}^{10}{{C}_{10}}$ways. Each entry will be written win, draw or loss. If we have one correct guess, 2 would be the number of incorrect guesses.
Let us consider if we have exactly 5 correct entries. We can fill them with the correct answer in $1\times 1\times 1\times 1\times 1={{\left( 1 \right)}^{5}}$ way. We can fill the rest $10-5=5$ entries with any one of the rest 2 guesses in $2\times 2\times 2\times 2\times 2={{2}^{5}}$ ways. So number of prediction lists with exactly 5 correct answer is ${}^{10}{{C}_{5}}{{\left( 1 \right)}^{5}}{{\left( 2 \right)}^{5}}$.Similarly we use the rule of product and then rule of sum and have the total number of outcomes as,

& n\left( A \right)={}^{10}{{C}_{5}}{{\left( 1 \right)}^{5}}{{\left( 2 \right)}^{5}}+{}^{10}{{C}_{6}}{{\left( 1 \right)}^{6}}{{\left( 2 \right)}^{4}}+{}^{10}{{C}_{7}}{{\left( 1 \right)}^{7}}{{\left( 2 \right)}^{3}}+{}^{10}{{C}_{8}}{{\left( 1 \right)}^{8}}{{\left( 2 \right)}^{2}}+{}^{10}{{C}_{9}}{{\left( 1 \right)}^{9}}{{\left( 2 \right)}^{1}} \\\ & +{}^{10}{{C}_{10}}{{\left( 1 \right)}^{10}}{{\left( 2 \right)}^{0}} \\\ & \Rightarrow n\left( A \right)=252\times 32+210\times 16+120\times 8+45\times 4+10\times 2+1=12585 \\\ \end{aligned}$$ So the probability that one entry contains at least 5 correct answers is $$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{12585}{{{3}^{10}}}$$ So the correct option is A. **So, the correct answer is “Option A”.** **Note:** We can alternatively find the answer by calculating the probability of a correct entry $\dfrac{1}{3}$ and probability of an correct entry $\dfrac{2}{3}$ and then use formula for conditional probability after selecting any 5 matches. If there are $m$ ways to one thing and $n$ ways to another thing then by rule of product we can do both simultaneously in $m\times n$ ways and by rule of sum we can do either of the thing in $m+n$ ways.