3x = x / 2
Mmc (2,1) = 2
6x = x
6x- x = 0
5x= 0
x = 0/5
x = 0
What the heck are you saying? I think you need you fix the question
Answer:
Definitely not a
Step-by-step explanation:
On “edge” it’s not a. I got it wrong and I failed with a 60
Answer:
80/81
Step-by-step explanation:
If a head is twice as likely to occur as a tail, then the probability of getting heads is 2/3 and the probability of getting tails is 1/3.
The probability of getting at least 1 head involves 4 scenarios:
1) 1 Head and 3 Tails
2) 2 Heads and 2 Tails
3) 3 Heads and 1 Tail
4) 4 Heads
Instead of calculate all these scenarios, you could calculate the opposite scenario: 4 Tails. The sum of all possible scenarios is 1, so:
P(at least one head) + P(no heads) = 1
Then, P(at least one head) = 1 - P(no heads)
The probability of 4 tails is:
P(no heads) = P(TTTT) = (1/3)(1/3)(1/3)(1/3)=1/81
Then, P(at least one head) = 1 - 1/81=80/81
