Atv set contains five circuit boards of type a, five of type b, and four of type c. the probability of failing in its first 5000 hours of use is 0.03 for a type a circuit board, 0.04 for a type b circuit board, and 0.03 for a type c circuit board. assuming that the failures of the various circuit boards are independent of one another, compute the probability that no circuit board fails in the first 5000 hours of use.

0.903264

Step-by-step explanation:

Given that a TV set contains five circuit boards of type A, five of type B, and four of type C. The probability of failing in its first 5000 hours of use is 0.03 for a type A circuit board, 0.04 for a type B circuit board, and 0.03 for a type C circuit board.

Let A' = the event that A fails, B' =  B fails and C'= C fails.

Probability that no circuit board fails = Prob (A'B'C')

= P(A')P(B')P(C')  (since A,B, C are independent, their complements are independent

= (1-0.03)(1-0.04)(1-0.03)

= 0.903264

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}