In a class of students 1/3 are left handed and 1/8 wear a dental brace.
A student is selected at random.
What is the probability of that student.
a) is left handed AND wears a brace
b) is left handed but does not wear a brace
c) does not wear a brace AND is not left handed
Note: 1/3 and 1/8 DOES NOT = 1 :/
Any ideas? for a) I did : 1/3 * 1/8
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- Thread Starter
- 12-10-2017 22:04
- 13-10-2017 14:49
The probability of a student not wearing a brace is 7/8 (1-1/8). Therefore the probability of a student being left handed and not wearing a brace is 1/3*7/8. Part c involves the same method.