Questions on Binomial Probability Models. Binomial probability answered by real tutors!

(a)  P(a six) = P(a six of 6) = 1/6.


(b)  P(an even number) = P(2 or 4 or 6 of 6) = 3/6 = 1/2.


(c)  P(a number greater than 3) = P(4 or 5 or 6 of six) = 3/6 = 1/2.


(d)  P(a three or a five) = P(3 or 5 of 6) = 2/6 = 1/3.

The deck = {B1,B2,B3,B4,Y1,Y2,Y3,Y4,G1,G2,G3,G4,R1,R2,R3,R4}

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(a)
I'll do B and R in the (a) part for you: 

B = {B1,B2,B3,B4}
R = {R1,R2,R3,R4}

You do the N part.  Here's how:
N = ?. If the number on the card is at most two, then that number is either 2,3,
or 4. So look through the deck and list all the cards that have a 2,3, or 4 on
them regardless of what letter it has before the number. List them all between
braces {} with commas separating them.

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(b) 
B ∩ R = ?.  Are there any blue cards that are also red cards? If so list
them between braces {} with commas separating them.  If not, write the empty set
∅  or { },

B ∪ R = ?.  List all the cards that are either in B and/or in R between braces
{}, with commas separating them. 

B ∩ N = ?.  Are there any cards that are both in B and N? If so, list them
between braces {} with commas separating them.  If not, write the empty set
∅  or { },

R ∪ N = ?.  List all the cards that are either in R and/or in N between braces
{}, with commas separating them. 

B = ?.  Just copy what I gave you above for B.

B ∪ R = ?.  List all the cards that are either in B and/or in R between braces
{}, with commas separating them.

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(c) In each of these, count the number of cards in the set which the probability
is asked for.  Write this as the numerator of a fraction. Then count the number
of cards in the deck, and write this as the denominator of the fraction. Reduce 
the fraction if possible.  This will be the answer in each case. 

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(d) Sets are said to be "mutually exclusive" when they have NO elements in
common.  [The word "exclusive" is not to be taken as what "exclusive" usually
means. When you see the words "mutually exclusive", think of the word
"exclusive" and though it were the word "excluding" instead. It means that each
of the two sets 'excludes' all (does not contain any) of the members or elements
of the other set.]

So use that to answer (d).

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(3)  Looks like you copied and pasted too much. LOL

Edwin

It is a binomial experiment.


There are 11 potential passengers. The probability that every given passenger will show up is 0.9.

The events for individual passengers are INDEPENDENT.

The question is: find the probability that of 11 passengers that bought tickets, at least 10 will show up
(it is the event "the flight is overbooked").


n = 11 (the number of trials);

p = 0.9  (the probability of "success" to each individual trial);

k >= 10  (the event of overbooking flight).


P(n=11; p=0.9; k>=10) = P(n=11; p=0.9; k=10) + P(n=11; p=0.9; k=11) =  +  =

   =  = 0.697  (rounded).    ANSWER

For each question, the probability to answer correctly is   = 0.8;
the probability to answer incorrectly is   = 0.2.


Since this experiment is binomial, the probability under the problem's question is

    P(4 of 20) =  =  = 1.30057E-08.  


ANSWER.  P(4 of 20) = 1.30057E-08.