A number from 1 to 10 is chosen at random. what is the probability of choosing an even number

probability of (a or b) is equal to probability of (a) + probability of (b) - probability of (a and b). in your problem: a is equal to choosing a number less than 5. b is equal to choosing an even number. there are 10 numbers to choose from. there are 4 numbers less than 5 (1,2,3,4), so the probability of choosing a number less than 5 is equal to 4/10. we get probability of (a) is equal to 4/10. there are 5 even numbers (2,4,6,8,10), so the probability of choosing an even number is 5/10. we get probability of (b) is equal to 5/10. there are 2 numbers that are both less than 5 and are both even (2,4), so the probability of choosing a number that is less than 5 and even at the same time is equal to 2/10. we get probability of (a and b) is equal to 2/10. the formula we are working from is, once again: probability of (a or b) is equal to probability of (a) + probability of (b) - probability of (a and b). this formula becomes 4/10 + 5/10 - 2/10 which is equal to 7/10. let's take a look and see if we're right. the numbers are: 1,2,3,4,5,6,7,8,9,10 the even numbers are 2,4,6,8,10 the numbers less than 5 are 1,2,3,4. if we just add up the numbers, we get a total of 9 numbers. 2 of them, however, are the same. those are the numbers 2 and 4. to avoid double counting them, we subtract them from the total. the total becomes 9 - 2 = 7 the numbers that are even or less than 5 are: 1,2,3,4,6,8,10 that's 7 out of the 10. the formula works. the whole rationale behind subtracting the probability of (a and b) is to remove the double counting since those numbers are being counted twice, once as members of the even number set and once as members of the number less than 5 set.

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