Question

Can someone check one Elementary Statistics question?

Answer 1

It has been estimated that 40% of marriages end in divorce. If you randomly select 9 recently married couples what is the probability that at least one of the marriages will end in divorce?

Answer 2

Im getting 0.0101 but someone else said its 0.9899

Answer 3

What is the point of posting that?

Answer 4

9C0 (0.4)^0 (0.6)^9

=1(0.4)^0 (0.6)^9 = 0.010077696 = 0.0101

Answer 5

so from this given details you know that 40% from marriage ends in divorce

what mean this exactly ?

that from 100 marriage 40 ends in divorce ,yes ?

Answer 6

using the rule of 3 simple you can writing

from 100 ends in divorce 40
from 9 --------------- x
---------------------------
x = 9*40/100 = ? %

hope helped easy

Answer 7

??

Answer 8

using this formula you get that from 9 marriage how many will ends in divorce - using this details above wrote ,given in this exercise

Answer 9

What formula? Sorry but none of my options are a percentage

Answer 10

Whatever you just did is completely incorrect.

Answer 11

options:

a. 0.0101
b. 0.9899
c. 0.9744
d. 0.1296

Answer 12

using these above wrote you get what is the percentage that from 9 marriage will ends in divorce

but you need getting what is the probability that the least will ends in divorce - yes ?

Answer 13

Sure?

Answer 14

C would be what I'd go with.

Answer 15

Can you show me how you came up with that??

Answer 16

I would do
1 - Pr(0 divorces)
that will be the probability of 1 or more divorces
in other words, the answer should be
1- 0.6^9

Answer 17

Thats what I originally did. But I was told to do this formula:

P(x)- nCxP^X q^n-x

Answer 18

Which still came out to the same answer for me.

Answer 19

P(x) = **

Answer 20

***
9C0 (0.4)^0 (0.6)^9 =1(0.4)^0 (0.6)^9 = 0.010077696 = 0.0101
***
that is the chance that all 9 will not divorce
1-0.0101 = 0.9899 is the chance that 1 or more divorces will occur out of the 9

Answer 21

Okay so thats what I was missing. Thank you

Answer 22

you could also do
P(1) + P(2) + ....+P(9)
but that is a lot of work

on the other hand we know out of 9, we must have either 0, 1, 2... 9 divorces
i.e. the sum of the probabilities must add up to 1
P(0) + P(1) + ...+P(9) = 1
and
P(1)+...+P(9) = 1 - P(0)

Answer 23

That makes sense. Thanks for explaining.

Answer 24

and the last thing is to be able to interpret Probability of 1 or more divorces
to mean P(1)+P(2)+...P(9)

Answer 25

and of course, know that "at least one" means (in this case) 1,2,3...9