For this case we must simplify the following expression:
So, if we apply distributive property to the terms within parentheses we have:
We simplify taking into account that:
Answer:
Answer:
(a) 0.343
(b) 0.657
(c) 0.189
(d) 0.216
(e) 0.353
Step-by-step explanation:
Let P(a vehicle passing the test) = p
Let P(a vehicle not passing the test) = q
q = 1 - p
q = 1 - 0.7 = 0.3
(a) P(all of the next three vehicles inspected pass) = P(ppp)
= 0.7 × 0.7 × 0.7
= 0.343
(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)
= (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)
= 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027
= 0.657
(c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)
= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)
= 0.063 + 0.063 + 0.063
= 0.189
(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)
= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)
= 0.063 + 0.063 + 0.063 + 0.027
= 0.216
(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?
P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)
= (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)
= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063
= 0.973
With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,
= 0.353
Answer:
3(x-2)(x+5)
Step-by-step explanation:
1. Factor out common term 3
3(x^2 +3x-10)
2. Factor (x^2 +3x-10)
3(x-2)(x+5)
