Question

A prime die uses the first six prime numbers on its faces, one prime per face. When two prime dice are rolled randomly, there is a 1 36 probability that the product of the two numbers will be equal to 4. What is the probability that the product has a units digit of 1?

Answer 1

Explanation:

Given

dice with Prime number is rolled

The number on the face of dice are 2,3,5,7,11 and 13

The sample space for the two dice.

`\left|\begin{array}c&2&3&5&7&11&13\\2&(2,2)&(2,3)&(2,5)&(2,7)&(2,11)&(2,13)\\3&(3,2)&(3,3)&(3,5)&(3,7)&(3,11)&(3,13)\\5&(5,2)&(5,3)&(5,5)&(5,7)&(5,11)&(5,13)\\7&(7,2)&(7,3)&(7,5)&(7,7)&(7,11)&(7,13)\\11&(11,2)&(11,3)&(11,5)&(11,7)&(11,11)&(11,13)\\13&(13,2)&(13,3)&(13,5)&(13,7)&(13,11)&(13,13)\end{array}\right|`

Their corresponding product is given below:

`\left|\begin{array}cccccc&2&3&5&7&11&13\\2&4&6&10&14&22&26\\3&6&9&15&21&33&39\\5&10&15&25&35&55&65\\7&14&21&35&49&77&91\\11&22&33&55&77&121&143\\13&26&39&65&91&143&169\end{array}\right|`

Now probability of getting product has a unit digit one is possible when outcome is

`(3,7),(7,3),(13,7),(7,13),(11,11)`

i.e. there are 5 favorable outcomes

So probability is
`(5)/(36)`