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A store is giving every customer who enters the store a scratch-off card labeled with numbers from 1 to 10.It is equally likely that any of the numbers from 1 to 10 will be labeled on a given card. If the card is an even number, the customer gets a 15 o/o discount on a purchase. If the card is an odd number greater than 6, the customer gets a 30 o/o discount. i will mark brainlest

A Store Is Giving Every Customer Who Enters The Store A Scratchoff Card Labeled With Numbers From 1 To 10It Is Equally Likely That Any Of The Numbers From 1 To class=

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Even number 15% off: 5/10 or 1/2 odds
odd number greater than 6 -30% off odds: 2/10 or 1/5
Otherwise 20% off: 3/10 chance/odds
We have 10 numbers. If a customer gets an even number, they will get a 15 percent discount. We know that there are 5 odd and 5 even numbers, therefore there is a 5/10 (50 percent) probability a customer gets a 15 percent discount. We need to find out how many odd numbers are greater than 6. 7 and 9. There are two odd numbers greater than 6, so there is a 2/10 (20 percent) probability a customer gets a 30 percent discount. We also need to find the probability if a customer doesn’t meet any of that criteria. If we add 5/10 + 2/10, we will get 7/10. 10/10 - 7/10 = 3/10 (30 percent).
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