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3 years ago

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Question 7 A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome.

Give the sample space describing all possible outcomes. Then give all of the outcomes for the event of rolling an odd number. If there is more than one element in the set, separate them with commas.

1 answer:

QveST [7]3 years ago

6 0

Answer:

1) $S=\{1,2,3,4,5,6\}$

2) $O=\{1,3,5\}$

Step-by-step explanation:

Given : A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome.

To find : Give the sample space describing all possible outcomes. Then give all of the outcomes for the event of rolling an odd number. If there is more than one element in the set, separate them with commas ?

Solution :

A number cube with faces labeled from 1 to 6 will be rolled once.

The sample space describing all possible outcomes is given by

$S=\{1,2,3,4,5,6\}$

All of the outcomes for the event of rolling an odd number is given by

$O=\{1,3,5\}$

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Answer:

Step-by-step explanation:

Let largest value of data be L

and smallest value of data be S.

We are given that the range and coefficient of range of the data are 17 and 0.2 respectively.

Range formula is given by = Largest value - Smallest value

17 = L - S

L = 17 + S ---------- [Equation 1]

Coefficient of range formula = (Largest - Smallest) ÷ (Largest + Smallest)

0.2 = l-s/l+s

0.2 = 17/l+s

L + S = 17/0.2 = 85

So, L + S = 85

17 + S + S = 85 {using equation 1}

2*S = 80

S = 80/2 = 40

Putting value of S in equation 1, we get L = 20 + 40 = 60

Therefore, Largest value, L = 60

smallest value, S = 40

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Step-by-step explanation:

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3 years ago

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

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2 years ago