Mathematics, 21.02.2020 03:52, madim1275

A sprinkler system inside an officebuilding has 2 types of not-very-reliable activation devices, A1 and A2, which operate independently. In case of a fire, the probability that device A1operates correctly is .60, and the probability that A2operates correctly is .30. If a fire breaks out, find theprobability that:

a. both devices operate correctly.
b. At least one device operates correctly.
c. Exactly one device fails to operate correctly.
d. Neither device operates correctly.
e. No more than one device operates correctly.

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answered: Crxymia

(a) The probability that both sprinklers operate correctly is 0.18.

(b) The probability that at least one sprinkler operates correctly is 0.72.

(d) The probability that neither sprinkler operates correctly is 0.28.

(e) The probability that no more than one sprinkler operates correctly is 0.82.

Step-by-step explanation:

The probability of the two sprinklers operating is:

P (A₁) = 0.60

P (A₂) = 0.30

The two sprinklers operate independently.

(a)

If events X and Y are independent than, $P(X\cap Y)=P(X)\times P(Y)$ .

Compute the probability that both sprinklers operate correctly as follows:

$P(A_{1}\cap A_{2})=P(A_{1})\times P(A_{2}) = 0.60 \times0.30=0.18$

Thus, the probability that both sprinklers operate correctly is 0.18.

(b)

Compute the probability that at least one sprinkler operates correctly as follows:

P (At least 1 operates correctly) = 1 - P (None operates correctly)

$=1-P(A^{c}_{1}\cap A^{c}_{2} )\\= 1 - P(A^{c}_{1})P(A^{c}_{2})\\= 1 -[1-P(A_{1})][1-P(A_{2})]\\= 1 - [1-0.60][1-0.30]\\=1-(0.40\times0.70)\\=0.72$

Thus, the probability that at least one sprinkler operates correctly is 0.72.

(c)

Compute the probability that exactly one sprinkler fails to operate as follows:

P (Exactly One fails to operates) = P (A₁ works but A₂ does not) +

P (A₁ does not works but A₂ works)

$=P(A_{1}\cap A^{c}_{2})+P(A^{c}_{1}\cap A_{2})\\=[P(A_{1})P(A^{c}_{2})]+[P(A^{c}_{1})P(A_{2})]\\=[P(A_{1})(1-P(A_{2}))]+[(1-P(A_{1}))P(A_{2})]\\=[0.60(1-0.30)]+[(1-0.60)0.30]\\=0.54$

Thus, the probability that exactly one sprinkler fails to operate is 0.54.

(d)

Compute the probability that neither sprinkler operates correctly as follows:

P (Neither operate correctly) = P (A₁ and A₂ does not works)

$=P(A^{c}_{1}\cap A^{c}_{2} )\\= P(A^{c}_{1})P(A^{c}_{2})\\= [1-P(A_{1})][1-P(A_{2})]\\= [1-0.60][1-0.30]\\=(0.40\times0.70)\\=0.28$

Thus, the probability that neither sprinkler operates correctly is 0.28.

(e)

Compute the probability that no more than one sprinkler operates correctly as follows:

P (No more than 1 operates correctly) = 1 - P (Both operates correctly)

$=1-P(A_{1}\cap A_{2})\\= 1-0.18\\=0.82$

Thus, the probability that no more than one sprinkler operates correctly is 0.82.

answered: Guest

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answered: Guest

it's both, now can you answer my most recent question?

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