The probability of rain on any particular day is 0.2. However, the probability of rain on a day after a rainy day is 0.85, whereas the probability of rain on a day after a non-rainy day is 0.1. On two consecutive days, find the probability of having exactly one rainy day

a)

i. = 0.17

ii. = 0.11

iii = 0.83

b)

i. = 0.1445

ii. = 0.0965

iii. = 0.241

step-by-step explanation:

to solve this problem we can represent the probabilities in a probability tree.

where r represents a rainy day and nr represents a day without rain

so, for part a) we have:

observe the attached diagram

then for part b) we have three consecutive days:

observe the attached diagram

a)

i. = 0.17

ii. = 0.11

iii = 0.83

b)

i. = 0.1445

ii. = 0.0965

iii. = 0.241

step-by-step explanation:

to solve this problem we can represent the probabilities in a probability tree.

where r represents a rainy day and nr represents a day without rain

so, for part a) we have:

observe the attached diagram

then for part b) we have three consecutive days:

observe the attached diagram

Cost of 7 gallons=$24.50

Cost of 1 gallon=24.50/7=3.5

Cost of 15 gallons=15*3.5=52.5

Cost of 15 gallons will be $52.5

The total nom of  code that can be used is equal to 5+3 = 8

Speed=Distance/time

Here,

distance=15m

time=1sec

speed=15/1=15m/sec

Distance=Speed*time

time=15min=15*60sec=900sec

Distance travelled in 15 min=15*900=13,500m

=13500/1000 km=13.5Km

Salesperson will make 6% of 1800

=(6/100)*1800

=108

Salesperson will make $108 in $1800 sales

The answer is in the image 

The answer is in the image 

30% of 420 are first year students

first year students = (30/100 )*420

=126 students

Student who are not in first year = 420-126

=294 students  

Let the father's age be x and son's be y

10 years before-

Father age=x-10

sons age=y-10

Given,

x-10=10(y-10)

x-10=10y-100

Given present age of father=40

therefore,

x=40

40-10=10y-100

10y-100=30

10y=130

y=130/10

y=13

Therefore present age of son=13years