You have cards numbered 1-10 and a number cube what are the odds of getting two prime numbers? The book says 1\5 but how do they get it.

I assume you pick a card, then throw the die

primes from 1-10 : 2,3,5,7 , so 4 of them
primes on a die: 2 3 5, so 3 of them
prob(2 primes) = (4/10)(3/6) = 1/5
prob(not 2 primes) = 4/5

odds in favour of getting two primes = (1/5) : (4/5
= 1 : 4

To calculate the odds of getting two prime numbers from a set of cards numbered 1-10 and a number cube, we need to determine the total number of possible outcomes and the number of favorable outcomes.

First, let's determine the total number of possible outcomes. You have a number cube, which has six sides numbered 1-6, and cards numbered 1-10. Since there are 6 possible outcomes from the cube and 10 possible outcomes from the cards, the total number of possible outcomes is 6 * 10 = 60.

Next, we need to find the number of favorable outcomes. A prime number is a number greater than 1 that is divisible only by 1 and itself. From the given cards, the prime numbers are 2, 3, 5, and 7. Therefore, the possible pairs of prime numbers are:

(2, 2), (2, 3), (2, 5), (2, 7)
(3, 2), (3, 3), (3, 5), (3, 7)
(5, 2), (5, 3), (5, 5), (5, 7)
(7, 2), (7, 3), (7, 5), (7, 7)

There are 16 possible pairs of prime numbers.

Therefore, the odds of getting two prime numbers are 16 favorable outcomes out of 60 possible outcomes. To simplify this fraction, we can divide both the numerator and denominator by their greatest common divisor, which is 4 in this case.

16/60 can be simplified to 4/15.

So, the correct answer is 4/15 or approximately 0.267 (rounded to three decimal places), not 1/5 as mentioned in the book.