Mathematics, 10.01.2020 02:31 cragajjar2

Events d and e are independent, with p(d)- 0.6 and p(d and e) - 0.18. which of the following is true? a. p(e)- 0.12 b. p(e) = 0.4 c. p(d or e)-0.28 d. p(d or e) 0.72 e. p(d or e)-0.9 16, mr. steckler travels through two intersections with traffic lights as he drives to the mercado. the traffic lights operate independently. the probability that both lights will be red when he reaches them is 0.22 the probability that the first light will be red and the second light will not be red is 0.33. what is the probability that the second light will be red when he reaches it? a. 0.40 b, 0.45 c. 0.50 d. 0.55 e. 0.60

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Answer from: lisaboden1701

Step-by-step explanation:

Given that Events D and E are independent, with P(D)- 0.6 and P(D and E) - 0.18.

P(D and E) = P(D)*P(E) (since independent)

Or P(E) = 0.3

Using this we find

P(Dor E) = 0.6+0.3-0.18 =0.72

QNo2

Since lights are independent

P(Both) = P(A) *P(B) =0.22 where A I light on and B second light on

P(AB') = P(A) P(B') = 0.33

When we add we get P(A) = 0.55

P(B) = $\frac{P(AB)}{P(A)} =0.6$

the probability that the second light will be red when he reaches it

= P(B) = 0.60

Answer from: alexius6608

Probabilities are used to determine the chances of an event.

The true statement is: $\mathbf{P(D\ or\ E) = 0.72}$ The probability that the second light will be red when he reaches it is 0.40

(a) Events D and E

The given parameters are:

$\mathbf{P(D) = 0.6}$

$\mathbf{P(D\ and\ E) = 0.18}$

Since the events are independent, then:

$\mathbf{P(D\ and\ E) = P(D) \times P(E)}$

Substitute known values

$\mathbf{0.18 = 0.6 \times P(E)}$

Divide both sides by 0.6

$\mathbf{0.3 = P(E)}$

Rewrite as:

$\mathbf{P(E) = 0.3 }$

Also:

$\mathbf{P(D\ or\ E) = P(D) + P(E) - P(D\ and\ E)}$

Substitute known values

$\mathbf{P(D\ or\ E) = 0.6 + 0.3 - 0.18}$

$\mathbf{P(D\ or\ E) = 0.72}$

So, the true statement is: $\mathbf{P(D\ or\ E) = 0.72}$

(b) Traffic lights

Let A and B represent the lights being red

The given parameters are:

$\mathbf{P(Both\ Red) = 0.22}$

$\mathbf{P(AB' ) = 0.33}$

Add both equations

$\mathbf{P(Both\ Red) +P(AB' ) = 0.22 + 0.33}$

$\mathbf{P(Both\ Red) +P(AB' ) = 0.55}$

Rewrite as:

$\mathbf{P(AB) +P(AB' ) = 0.55}$

Express as

$\mathbf{P(A) \times P(B) +P(A) \times P(B' ) = 0.55}$

Factor out P(A)

$\mathbf{P(A) \times [P(B) + P(B' )] = 0.55}$

Opposite probabilities add up to 1

$\mathbf{P(A) \times1 = 0.55}$

$\mathbf{P(A) = 0.55}$

Recall that: $\mathbf{P(Both\ Red) = 0.22}$

So, we have:

$\mathbf{P(A) \times P(B) = 0.22}$

Substitute $\mathbf{P(A) = 0.55}$

$\mathbf{0.55 \times P(B) = 0.22}$

Divide both sides by 0.55

$\mathbf{P(B) = 0.40}$

Hence, the probability that the second light will be red when he reaches it is 0.40

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Events d and e are independent, with p(d)- 0.6 and p(d and e) - 0.18. which of the following is true...

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