Let’s first see the probability of landing an even number. We have 6 sides, and 3 are labeled with an even number. So our chances of rolling a number cube are 3/6 (or 1/2 when simplified).
Next, the probability of rolling a number less then five is 4/6 (or 2/3 simplified), since we have 4 sides labeled with a number less than 5.
To compute two or more probabilities, we multiply them. In this case, the fractions already share the same denominator, 6 (because it’s a cube with 6 sides). So we get:
3/6 * 4/6
We then square 6, which is 36, and multiply 3 by 4, which is 12. This gives us 12/36, which is 1/3 in simplified form.
So there’s out answer: 1/3.
Complete question :
Suppose someone gives you 8 to 2 odds that you cannot roll two even numbers with the roll of two fair dice. This means you win $8 if you succeed and you lose $2 if you fail. What is the expected value of this game to you? What can you expect if you play 100 times.
Answer:
$0.5 ; win $50 with 100 rolls
Step-by-step explanation:
From a roll of two fair dice; probability of obtaining an even number :
Even numbers = (2, 4, 6) = 3
P = 3 /6 = 1 /2
For 2 fair dice ; probability of rolling two even numbers : independent event.
1/2 * 1/2 = 1/4
Hence, p(success) = 1/4 ; P(failure) = 1 - 1/4 = 3/4
Probability table
Winning = $8 or loss = - $2
X : ____ 8 ______ - 2
P(x) __ 1/4 ______ 3/4
Expected value : E(x) = ΣX*P(x)
E(x) = (8 * 1/4) + (-2 * 3/4)
E(x) = 2 - 1.5
E(x) = $0.5
Since expected value is positive, the expect to win
If played 100 times;
Expected value = 100 * $0.5 = $50
Answer:
x= -4.284 and y= 1.432
Step-by-step explanation:
Answer:
(a)
The probability that you stop at the fifth flip would be
(b)
The expected numbers of flips needed would be
Therefore, suppose that 
, then the expected number of flips needed would be 1/0.5 = 2.
Step-by-step explanation:
(a)
Case 1
Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be
Case 2
Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be
Therefore the probability that you stop at the fifth flip would be
(b)
The expected numbers of flips needed would be
Therefore, suppose that , then the expected number of flips needed would be 1/0.5 = 2.
It’s true!! Hope this helped!!!!
