Math / Data & Statistics

QuestionLet $A, B$ be two independent events of a sample space, where $P(A)=0.4$ , $P(\bar{B})=0.6$ . Then $P(\bar{A} \cup B)=$ 0.8 0.2 0.76 0.18 None of these

Studdy Solution

STEP 1

What is this asking? What's the probability of not A happening or B happening, if A and B are independent? Watch out! Don't mix up "and" and "or," and remember how independence works!

STEP 2

1. Find P(B)
2. Find P(not A)
3. Apply the formula for P(not A or B)

STEP 3

We're given $P(\bar{B}) = 0.6$ , which means the probability of B not happening is 0.6.
Since B either happens or it doesn't, the probabilities must add to one!
So, $P(B) = 1 - P(\bar{B})$ .

STEP 4

Let's calculate $P(B)$ : $P(B) = 1 - 0.6 = 0.4$ So, the probability of B happening is 0.4!

STEP 5

We're given $P(A) = 0.4$ .
Similar to what we did before, since A either happens or it doesn't, we know $P(\bar{A}) = 1 - P(A)$ .

STEP 6

Let's calculate $P(\bar{A})$ : $P(\bar{A}) = 1 - 0.4 = 0.6$ So, the probability of A not happening is 0.6!

STEP 7

Since A and B are independent, we can use the formula: $P(\bar{A} \cup B) = P(\bar{A}) + P(B) - P(\bar{A} \cap B)$ , where $P(\bar{A} \cap B) = P(\bar{A}) \cdot P(B)$ .
This formula tells us the probability of not A happening or B happening (or both!). The last term accounts for the overlap, so we don't double-count when both events happen.

STEP 8

First, let's calculate the probability of both not A and B happening: $P(\bar{A} \cap B) = P(\bar{A}) \cdot P(B) = 0.6 \cdot 0.4 = 0.24$

STEP 9

Now, let's plug everything into our formula: $P(\bar{A} \cup B) = P(\bar{A}) + P(B) - P(\bar{A} \cap B) = 0.6 + 0.4 - 0.24$

STEP 10

Finally, let's calculate the result: $P(\bar{A} \cup B) = 0.6 + 0.4 - 0.24 = 1 - 0.24 = 0.76$

STEP 11

The probability of not A happening or B happening is 0.76!

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QuestionLet $A, B$ be two independent events of a sample space, where $P(A)=0.4$ , $P(\bar{B})=0.6$ . Then $P(\bar{A} \cup B)=$ 0.8 0.2 0.76 0.18 None of these

Studdy Solution

STEP 1

What is this asking? What's the probability of not A happening or B happening, if A and B are independent? Watch out! Don't mix up "and" and "or," and remember how independence works!

STEP 2

1. Find P(B)
2. Find P(not A)
3. Apply the formula for P(not A or B)

STEP 3

We're given $P(\bar{B}) = 0.6$ , which means the probability of B not happening is 0.6.
Since B either happens or it doesn't, the probabilities must add to one!
So, $P(B) = 1 - P(\bar{B})$ .

STEP 4

Let's calculate $P(B)$ : $P(B) = 1 - 0.6 = 0.4$ So, the probability of B happening is 0.4!

STEP 5

We're given $P(A) = 0.4$ .
Similar to what we did before, since A either happens or it doesn't, we know $P(\bar{A}) = 1 - P(A)$ .

STEP 6

Let's calculate $P(\bar{A})$ : $P(\bar{A}) = 1 - 0.4 = 0.6$ So, the probability of A not happening is 0.6!

STEP 7

STEP 8

First, let's calculate the probability of both not A and B happening: $P(\bar{A} \cap B) = P(\bar{A}) \cdot P(B) = 0.6 \cdot 0.4 = 0.24$

STEP 9

Now, let's plug everything into our formula: $P(\bar{A} \cup B) = P(\bar{A}) + P(B) - P(\bar{A} \cap B) = 0.6 + 0.4 - 0.24$

STEP 10

Finally, let's calculate the result: $P(\bar{A} \cup B) = 0.6 + 0.4 - 0.24 = 1 - 0.24 = 0.76$

STEP 11

The probability of not A happening or B happening is 0.76!

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QuestionLet A,BA, BA,B be two independent events of a sample space, where P(A)=0.4P(A)=0.4P(A)=0.4, P(Bˉ)=0.6P(\bar{B})=0.6P(Bˉ)=0.6. Then P(Aˉ∪B)=P(\bar{A} \cup B)=P(Aˉ∪B)= 0.8 0.2 0.76 0.18 None of these

Studdy Solution

STEP 1

What is this asking? What's the probability of *not* A happening *or* B happening, if A and B are independent? Watch out! Don't mix up "and" and "or," and remember how independence works!

STEP 2

1. Find P(B)2. Find P(not A)3. Apply the formula for P(not A or B)

STEP 3

We're given P(Bˉ)=0.6P(\bar{B}) = 0.6P(Bˉ)=0.6, which means the probability of B *not* happening is **0.6**.Since B either happens or it doesn't, the probabilities must add to **one**!So, P(B)=1−P(Bˉ)P(B) = 1 - P(\bar{B})P(B)=1−P(Bˉ).

STEP 4

Let's **calculate** P(B)P(B)P(B): P(B)=1−0.6=0.4P(B) = 1 - 0.6 = 0.4P(B)=1−0.6=0.4 So, the probability of B happening is **0.4**!

STEP 5

We're given P(A)=0.4P(A) = 0.4P(A)=0.4.Similar to what we did before, since A either happens or it doesn't, we know P(Aˉ)=1−P(A)P(\bar{A}) = 1 - P(A)P(Aˉ)=1−P(A).

STEP 6

Let's **calculate** P(Aˉ)P(\bar{A})P(Aˉ): P(Aˉ)=1−0.4=0.6P(\bar{A}) = 1 - 0.4 = 0.6P(Aˉ)=1−0.4=0.6 So, the probability of A *not* happening is **0.6**!

STEP 7

STEP 8

First, let's **calculate** the probability of both *not* A and B happening: P(Aˉ∩B)=P(Aˉ)⋅P(B)=0.6⋅0.4=0.24P(\bar{A} \cap B) = P(\bar{A}) \cdot P(B) = 0.6 \cdot 0.4 = 0.24P(Aˉ∩B)=P(Aˉ)⋅P(B)=0.6⋅0.4=0.24

STEP 9

Now, let's **plug** everything into our formula: P(Aˉ∪B)=P(Aˉ)+P(B)−P(Aˉ∩B)=0.6+0.4−0.24P(\bar{A} \cup B) = P(\bar{A}) + P(B) - P(\bar{A} \cap B) = 0.6 + 0.4 - 0.24P(Aˉ∪B)=P(Aˉ)+P(B)−P(Aˉ∩B)=0.6+0.4−0.24

STEP 10

Finally, let's **calculate** the result: P(Aˉ∪B)=0.6+0.4−0.24=1−0.24=0.76P(\bar{A} \cup B) = 0.6 + 0.4 - 0.24 = 1 - 0.24 = 0.76P(Aˉ∪B)=0.6+0.4−0.24=1−0.24=0.76

STEP 11

The probability of *not* A happening *or* B happening is **0.76**!

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QuestionLet $A, B$ be two independent events of a sample space, where $P(A)=0.4$ , $P(\bar{B})=0.6$ . Then $P(\bar{A} \cup B)=$ 0.8 0.2 0.76 0.18 None of these

What is this asking? What's the probability of not A happening or B happening, if A and B are independent? Watch out! Don't mix up "and" and "or," and remember how independence works!

1. Find P(B)
2. Find P(not A)
3. Apply the formula for P(not A or B)

We're given $P(\bar{B}) = 0.6$ , which means the probability of B not happening is 0.6.
Since B either happens or it doesn't, the probabilities must add to one!
So, $P(B) = 1 - P(\bar{B})$ .

Let's calculate $P(B)$ : $P(B) = 1 - 0.6 = 0.4$ So, the probability of B happening is 0.4!

We're given $P(A) = 0.4$ .
Similar to what we did before, since A either happens or it doesn't, we know $P(\bar{A}) = 1 - P(A)$ .

Let's calculate $P(\bar{A})$ : $P(\bar{A}) = 1 - 0.4 = 0.6$ So, the probability of A not happening is 0.6!

First, let's calculate the probability of both not A and B happening: $P(\bar{A} \cap B) = P(\bar{A}) \cdot P(B) = 0.6 \cdot 0.4 = 0.24$

Now, let's plug everything into our formula: $P(\bar{A} \cup B) = P(\bar{A}) + P(B) - P(\bar{A} \cap B) = 0.6 + 0.4 - 0.24$

Finally, let's calculate the result: $P(\bar{A} \cup B) = 0.6 + 0.4 - 0.24 = 1 - 0.24 = 0.76$

The probability of not A happening or B happening is 0.76!