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?

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Brittany Patton · Accepted Answer

Of the ten integers between one and ten inclusive, six are more than four (that is, 5,6,7,8,9,10).The probability is calculated by dividing the number of likely outcomes by the total number of outcomes:&amp;nbsp;P(&amp;gt;4)=#of favorable outcomes&amp;nbsp;#&amp;nbsp;of possible outcomes=6105 of the 10 integers between 1 and 10 inclusive are odd (that is, 1,3,5,7,9). P(odd)=# of favorable outcomes/# of possible outcomes=510Three of the ten integers, ranging from one to ten, are more than four and odd (that is, 5,7,9).&amp;nbsp;P(&amp;gt;4&amp;nbsp;and odd)=#&amp;nbsp;of favorable outcomes/# of possible outcomes=310If there are any two occurrences, apply the general addition rule:&amp;nbsp;P(AUB)=P(A)+P(B)−P(A⋂B)P(&amp;gt;4 or odd)&amp;nbsp;=P(&amp;gt;4)+P(odd)−P(&amp;gt;4and odd)=610+510−310=6+5−310=810=45

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

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