.
In all, there are 36 possible different outcomes.
Of them, 6 outcomes are pairs with equal appearing, and the rest 30 pairs are with different appearing.
For our conditional probability problem, these 30 pairs comprise our reduced space of events.
The sum of 6 produce these five pairs (1,5), (2,4), (3,3), (4,2), (5,1).
Of these 5 pairs, we exclude the pair (3,3), since this pair is of equal appearance.
Thus the number of favorable pairs in our problem is 4.
The probability under the problem's question is
P = = = . ANSWER
Solved.
Solve part (b) using the same logic.
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In this site, there are two lessons for problems on conditional probability. They are
- Conditional probability problems
- Conditional probability problems REVISITED
Read them to develop your skills and knowledge.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Solved problems on Probability"
and "Additional problems on Probability".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.