Mathematics, 18.07.2019 15:30 luckye

Someone plz answer giving 45 points for this question! a fair decimal die is tossed five times. find the probability of obtaining an odd number of even digits.

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Well, an even decimal die has probability 0.5 of rolling even and 0.5 of rolling odd.

The chances of rolling exactly one even number on five rolls is P(even)⋅(P(odd))4=0.5⋅0.54=0.55=132." role="presentation" style="margin: 0px; padding: 0px; outline: 0px; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">P(even)⋅(P(odd))4=0.5⋅0.54=0.55=132.P(even)⋅(P(odd))4=0.5⋅0.54=0.55=132.

The chances of rolling exactly three even numbers on five rolls is (P(even))3⋅(P(odd))2=0.53⋅0.52=0.55=132" role="presentation" style="margin: 0px; padding: 0px; outline: 0px; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">(P(even))3⋅(P(odd))2=0.53⋅0.52=0.55=132(P(even))3⋅(P(odd))2=0.53⋅0.52=0.55=132.

The chances of rolling exactly five even numbers on five rolls is (P(even))5=0.55=132" role="presentation" style="margin: 0px; padding: 0px; outline: 0px; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">(P(even))5=0.55=132(P(even))5=0.55=132.

Further, the probability of independent events is the sum of their probabilities, so the probability of rolling an odd number of even numbers is 332" role="presentation" style="margin: 0px; padding: 0px; outline: 0px; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">332

Answer from: Quest

step-by-step explanation:

Answer from: Quest

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

=3x+1

=3(2)+1

since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

=x+1

=2+1

now, let's find g(2)g(2):

g(x)

g(2)

=2x

=2⋅2

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

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Someone plz answer giving 45 points for this question! a fair decimal die is tossed five times. fin...