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Hint: First use the given odd to find the probability of passing the first student and the probability of passing the second student. Then use them to find the probability that both the students pass the test.
Complete step-by-step answer:
It is given that the odds in the favor of one student passing a test are 3:7 and the odds against another student passing it are 3:5.
We have to find the probability that both the students pass the test.
As we know that the odds are defined as the probability of occurring the event divided by the probability of not occurring the event.
It is given that the odds in the favor of one student passing a test are 3:7, it means that the probability of passing the student is $\left( {\dfrac{3}{{3 + 7}} = \dfrac{3}{{10}}} \right)$ and the probability of not passing the student is$\left( {\dfrac{7}{{3 + 7}} = \dfrac{7}{{10}}} \right)$.
Hence, the probability of passing the first student is$\left( {\dfrac{3}{{10}}} \right)$.
It is also given that the odds against another student passing it are 3:5. It means that the probability of not passing another student is $\left( {\dfrac{3}{{3 + 5}} = \dfrac{3}{8}} \right)$ and the probability of passing the student is$\left( {\dfrac{5}{{3 + 5}} = \dfrac{5}{8}} \right)$.
Now, we have the probability of passing the first student is $\left( {\dfrac{3}{{10}}} \right)$ and the probability of passing the second student is$\left( {\dfrac{5}{8}} \right)$.
The probability of passing both the students is the product of the probability of passing the first student and the probability of passing the second student.
Thus, the probability of passing both the students is given as:
$ = {\text{Probabiity of passing }}{{\text{1}}^{st}}{\text{ student}} \times {\text{Probabiity of passing }}{{\text{2}}^{st}}{\text{ student}}$
$ = \dfrac{3}{{10}} \times \dfrac{5}{8}$
$ = \dfrac{3}{{16}}$
Hence, the probability of passing both students is$\dfrac{3}{{16}}$.
Note: It has to notice that the odds in the favor of one student passing a test are 3:7 and the odds is given in favor so the $\left( {\dfrac{3}{{10}} = 0.3} \right)$ is the probability of passing but in case of the odds against another student passing it are 3:5, then the probability of passing is$\left( {\dfrac{5}{8} = 0.625} \right)$.
Complete step-by-step answer:
It is given that the odds in the favor of one student passing a test are 3:7 and the odds against another student passing it are 3:5.
We have to find the probability that both the students pass the test.
As we know that the odds are defined as the probability of occurring the event divided by the probability of not occurring the event.
It is given that the odds in the favor of one student passing a test are 3:7, it means that the probability of passing the student is $\left( {\dfrac{3}{{3 + 7}} = \dfrac{3}{{10}}} \right)$ and the probability of not passing the student is$\left( {\dfrac{7}{{3 + 7}} = \dfrac{7}{{10}}} \right)$.
Hence, the probability of passing the first student is$\left( {\dfrac{3}{{10}}} \right)$.
It is also given that the odds against another student passing it are 3:5. It means that the probability of not passing another student is $\left( {\dfrac{3}{{3 + 5}} = \dfrac{3}{8}} \right)$ and the probability of passing the student is$\left( {\dfrac{5}{{3 + 5}} = \dfrac{5}{8}} \right)$.
Now, we have the probability of passing the first student is $\left( {\dfrac{3}{{10}}} \right)$ and the probability of passing the second student is$\left( {\dfrac{5}{8}} \right)$.
The probability of passing both the students is the product of the probability of passing the first student and the probability of passing the second student.
Thus, the probability of passing both the students is given as:
$ = {\text{Probabiity of passing }}{{\text{1}}^{st}}{\text{ student}} \times {\text{Probabiity of passing }}{{\text{2}}^{st}}{\text{ student}}$
$ = \dfrac{3}{{10}} \times \dfrac{5}{8}$
$ = \dfrac{3}{{16}}$
Hence, the probability of passing both students is$\dfrac{3}{{16}}$.
Note: It has to notice that the odds in the favor of one student passing a test are 3:7 and the odds is given in favor so the $\left( {\dfrac{3}{{10}} = 0.3} \right)$ is the probability of passing but in case of the odds against another student passing it are 3:5, then the probability of passing is$\left( {\dfrac{5}{8} = 0.625} \right)$.
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