Answer:
(a) 0.93
(b) 0.38
Step-by-step explanation:
A attends class 30% of the time, so DO NOT attend 70%
B attends class 50% of the time, so DO NOT attend 50%
C attends class 80% of the time, so DO NOT attend 20%
Writing as probabilities:
P(A) = 0.30 and P(A') = 0.70
P(B) = 0.50 and P(B') = 0.50
P(C) = 0.80 and P(C') = 0.20
(a) the probability that at least one of them will be in class on a particular day
Let's call Q the event of none of them be in class on a particular day
Probability of at least one be in class is the complement of none of them be there, so: 1 - P(Q)
P(Q) = 0.7*0.5*0.2 = 0.07
1 - P(Q) = 1 - 0.07 = 0.93
(b) the probability that exactly one of them will be in class on a particular day?
One of them exactly be in class is
A is B not C not or A not B is C not or A not B not and C is, so
P(A).P(B').P(C') + P(A').P(B).P(C') + P(A').P(B').P(C)
0.3*0.5*0.2 + 0.7*0.5*0.2 + 0.7*0.5*0.8 =
0.03 + 0.07 + 0.28 = 0.38
<u>Answer:</u>
The solution set of given equations -x-y-z = -8 and - 4x + 4y + 5z = 7 and 2x + 2z = 4 is (3, 6, -1)
<u>Solution:</u>
Given, set linear equations are
-x – y – z = -8 ⇒ x + y + z = 8 → (1)
-4x + 4y + 5z = 7 ⇒ 4x – 4y – 5z = -7 → (2)
2x + 2z = 4 ⇒ x + z = 2 → (3)
We have to solve the above given equations using substitution method.
Now take (3), x + z = 2 ⇒ x = 2 – z
So substitute x value in (1)
(1) ⇒ (2 – z) + y + z = 8 ⇒ 2 + y + z – z = 8 ⇒ y + 0 = 8 – 2 ⇒ y = 6.
Now substitute x and y values in (2)
(2) ⇒ 4(2 – z) – 4(6) – 5z = - 7 ⇒ 8 – 4z – 24 – 5z = -7 ⇒ -9z – 16 = -7 ⇒ 9z = 7 – 16 ⇒ 9z = -9 ⇒ z = -1
Now substitute z value in (3)
(3) ⇒ x – 1 = 2 ⇒ x = 2 + 1 ⇒ x = 3
Hence, the solution set of given equations is (3, 6, -1).
Let <span>Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be represented by the letters J, C, G, M, E, D, A, and S respectively. </span>
<span>In part IV we are asked:
</span><span>What is the sample space of the pairs of potential clients that could be chosen?
</span><span>
Since the Sample Space is the set of all possible outcomes, we need to make a set (a list) of all the possible pairs, which are as follows:
{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)
, </span>(C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
<span>
</span> , (G, M), (G, E), (G, D), (G, A), (G, S)
<span>
,</span>(M, E), (M, D), (M, A), (M, S)
<span>
, </span>(E, D), (E, A), (E, S) <span>
, </span>(D, A), (D, S)
, (A, S).}
We can check that the number of the elements of the sample space, n(S) is
1+2+3+4+5+6+7=28.
This gives us the answer to the first question: <span>How many pairs of potential clients can be randomly chosen from the pool of eight candidates?
(Answer: 28.)
II) </span><span>What is the probability of any particular pair being chosen?
</span>
The probability of a particular pair to be picked is 1/28, as there is only one way of choosing a particular pair, out of 28 possible pairs.
III) <span>What is the probability that the pair chosen is Jacob and Meg or Geraldo and Sally?
The probability of choosing (J, M) or (G, S) is 2 out of 28, that is 1/14.
Answers:
I) 28
II) 1/28</span>≈0.0357
III) 1/14≈0.0714
IV)
{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)
, (C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
, (G, M), (G, E), (G, D), (G, A), (G, S)
,(M, E), (M, D), (M, A), (M, S)
, (E, D), (E, A), (E, S)
, (D, A), (D, S)
, (A, S).}
