Hey! please help i’ll give brainliest
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Answer:
the probability the die chosen was green is 0.9
Step-by-step explanation:
Given that:
A bag contains two six-sided dice: one red, one green.
The red die has faces numbered 1, 2, 3, 4, 5, and 6.
The green die has faces numbered 1, 2, 3, 4, 4, and 4.
From above, the probability of obtaining 4 in a single throw of a fair die is:
P (4 | red dice) =
P (4 | green dice) = =
A die is selected at random and rolled four times.
As the die is selected randomly; the probability of the first die must be equal to the probability of the second die =
The probability of two 1's and two 4's in the first dice can be calculated as:
=
=
=
=
=
=
The probability of two 1's and two 4's in the second dice can be calculated as:
=
=
=
=
=
∴
The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )
The probability of two 1's and two 4's in both die =
The probability of two 1's and two 4's in both die =
The probability of two 1's and two 4's in both die =
By applying Bayes Theorem; the probability that the die was green can be calculated as:
P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)
P(second die (green) | two 1's and two 4's ) =
P(second die (green) | two 1's and two 4's ) =
P(second die (green) | two 1's and two 4's ) = 0.9
Thus; the probability the die chosen was green is 0.9
Answer:
Step-by-step explanation:
Hello!
The variables of interest are:
X₁: Number of cases of polio observed in kids that received the placebo vaccine.
n₁= 201299 total children studied
x₁= 110 cases observed
X₂: Number of cases of polio observed in kids that received the experimental vaccine.
n₂= 200745 total children studied
x₂= 33 cases observed
These two variables have a binomial distribution. The parameters of interest, the ones to compare, are the population proportions: p₁ vs p₂
You have to test if the population proportions of children who contracted polio in both groups are different: p₂ ≠ p₁
a)
H₀: p₂ = p₁
H₁: p₂ ≠ p₁
α: 0.05
Sample proportion placebo p'₁= x₁/n₁= 110/201299= 0.0005
Sample proportion vaccine p'₂= x₂/n₂= 33/200745= 0.0002
Pooled sample proportion p'= (x₁+x₂)/(n₁+n₂)= (110+33)/(201299+200745)= 0.0004
This test is two-tailed, using the critical value approach, you have to determine two critical values:
Then if ≤ -1.96 or if
≥ 1.96, the decision is to reject the null hypothesis.
If -1.96 < < 1.96, the decision is to not reject the null hypothesis.
⇒ = -4.76, the decision is to reject the null hypothesis.
b)
H₀: p₂ = p₁
H₁: p₂ ≠ p₁
α: 0.01
The value of = -4.76 doesn't change, since we are working with the same samples.
The only thing that changes alongside with the level of significance is the rejection region:
Then if ≤ -2.576or if
≥ 2.576, the decision is to reject the null hypothesis.
If -2.576< < 2.576, the decision is to not reject the null hypothesis.
⇒ = -4.76, the decision is to reject the null hypothesis.
c)
Remember the level of significance (probability of committing type I error) is the probability of rejecting a true null hypothesis. This means that the smaller this value is, the fewer chances you have of discarding the true null hypothesis. But as you know, you cannot just reduce this value to zero because, the smaller α is, the bigger β (probability of committing type II error) becomes.
Rejecting the null hypothesis using different values of α means that there is a high chance that you reached a correct decision (rejecting a false null hypothesis)
I hope this helps!