Answer:
Step-by-step 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.
[tex]\left|\begin{array}{c|cccccc}&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|[/tex]
Their corresponding product is given below:
[tex]\left|\begin{array}{c|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|[/tex]
Now probability of getting product has a unit digit one is possible when outcome is
[tex](3,7),(7,3),(13,7),(7,13),(11,11)[/tex]
i.e. there are 5 favorable outcomes
So probability is [tex]\frac{5}{36}[/tex]
