Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. What is the probability that the number will be more than 4 or odd

fortdefruitI 2021-06-28 Answered
Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. What is the probability that the number will be more than 4 or odd? (Enter your probability as a fraction.)

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Expert Answer

Arnold Odonnell
Answered 2021-06-29 Author has 8924 answers

6 of the 10 integers between 1 and 10 inclusive are more than 4 (that is, 5,6,7,8,9,10).
The probability is the number of favorable outcomes divided by the number of possible outcomes: \(P(>4)=\)\(\frac{\text{ of favorable outcomes}}{\text{ of possible outcomes}}\)=\(\displaystyle\frac{{6}}{{10}}\)
5 of the 10 integers between 1 and 10 inclusive are odd (that is, 1,3,5,7,9). P(odd)=\(\frac{\text{ of favorable outcomes}}{\text{ of possible outcomes}}\)=\(\displaystyle\frac{{5}}{{10}}\)
3 of the 10 integers between 1 and 10 inclusive are move than 4 and odd (that is, 5,7,9). \(P(>4\) and odd)=\(\frac{\text{ of favorable outcomes}}{\text{ of possible outcomes}}\)=\(\displaystyle\frac{{3}}{{10}}\)
Use the General addition rule for any two events: \(\displaystyle{P}{\left({A}{U}{B}\right)}={P}{\left({A}\right)}+{P}{\left({B}\right)}-{P}{\left({A}⋂{B}\right)}\)
\(P(>4\ \text{or odd}) =P(>4)+P(\text{odd})-P(>4\ \text{and odd}) =\displaystyle\frac{6}{{10}}+\frac{5}{{10}}-\frac{3}{{10}}=\frac{{{6}+{5}-{3}}}{{10}}=\frac{8}{{10}}=\frac{4}{{5}}\)

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