A standard cube has 6 possible outcomes. But since we want to find P(4+ or odd), the outcomes become more narrow.
So:
6 outcomes -> 1, 2, 3, 4, 5, 6
But we want to find 4+
So:
4+ = 4, 5, 6
We also want to find all odd possibilities.
So:
odd = 1, 3, 5
As you can see, there is a 5 in both the 4+ and the odd numbers so 5 will only be counted once.
P(4+ or odd) = 1, 3, 4, 5, 6
So the probability of having an outcome that is P(4+ or odd) is:
5/6 or 0.8333 or 83.33%
Hope I helped!
Multiply 14 to each term: 14a+70+84b
I think the question 23 is e.
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
There are 11 numbers on cards.
6 cards contain even numbers.
Answer: p(even) = 6/11
There is only one card with a 7. After the first card is picked and returned, there are 11 cards in the bag again.
p(even then 7) = p(even) * p(7) = 6/11 * 1/11 = 6/121
Answer: p(even then 7) = 6/121
Answer:
27
Step-by-step explanation:
