A standard number cube is tossed. find p(greater than 2 or even).
=4/6
=2/3
There are 4 numbers greater than 2, which also contains the two even numbers (4 and 6) so the set = {3, 4, 5, 6} which is four numbers of a total 6 so therefore the p = 4/6 which =2/3
Hence, we get:
\(P(\text{Greater\ than\ two\ or\ even})=\dfrac{5}{6}\)
Let A denote the event that the number is greater than 2.
and B denote the event that the number is even.
Let P denote the probability of an event.
Then we are asked to find:
P(A∪B)
We know that:
P(A∪B)=P(A)+P(B)-P(A∩B)
Now, we know that:
\(P(A)=\dfrac{4}{6}\)
( Since, there are a total of 4 numbers which are greater than 2 i.e. {3,4,5,6} )
Also,
\(P(B)=\dfrac{3}{6}\)
( Since there are a total of 3 numbers which are even.
i.e. {2,4,6} )
Also,
\(P(A\bigcap B)=\dfrac{2}{6}\)
( Since there are two elements which are both even as well as greater than 2 i.e. {4,6} )
Hence, we have:
\(P(A\bigcup B)=\dfrac{4}{6}+\dfrac{3}{6}-\dfrac{2}{6}\\\\i.e.\\\\P(A\bigcup B)=\dfrac{4+3-2}{6}\\\\i.e.\\\\P(A\bigcup B)=\dfrac{5}{6}\)