Hello Alexpavlos,
Let me try helping you with this one. This question can be solved in quite a few ways.
1) Let us simply state the possibilities. This method works for this problem due to the limited number of possibilities
Ways of getting at least a head and an even number =6
HT8
TH8
HT10
TH10
HH8
HH10
Ways of not getting at least one head and an even number=6
TT8
TT10
HT3
TH3
HH3
TT3
Total probability=6/12=1/2
Method 2)
Probability of getting at least one head and an even number= probability of getting head*probability of getting tail *probability of getting an even number+probability of getting tail*probability of getting head*probability of getting an even number+probability of getting head*probability of getting head*probability of getting an even number
=(1/2)*(1/2)*(2/3)+(1/2)*(1/2)*(2/3)+(1/2)*(1/2)*(2/3)=1/2
Hope this helps! Let me know if I can help you further.
alexpavlos wrote:
A fair coin is to be tossed twice and an integer is to be selected at random from one of the integers 3, 8, and 10. What is the probability that at least one head is tossed and an even integer is selected?
a) 1/6
b) 1/4
c) 1/3
d) 1/2
e) 7/12