A certain deck of cards contains 2 blue cards, 2 red cards

hello

I have computed the probability for both Cards to be blue and substract one
hence total card to 8
probability for the first card 2/8 OR 1/4
For the second 1/7
HENCE 1/28 FOR CARDS

1-1/28 = 27/28

HOPE this helps
best regards

@keiraria

Thanks for your help but can you solve this one for me too, its same as above only the language has been changed , its NONE now .

If it states that both are not blue then one can be blue , try this one below.


A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that NONE are blue?

hello Joy
thanks for asking me . in fact the way the question is asked has double meaning and it is
a little ambigious
it is noone are blue it is what I have done :27/28


but if it is that both can not be blue
that is the probability to have noNE blue + the probablity to have maximum one blue and hence one of any other colors

hence or case

27/28+ 2/8x6/7

Hope it is right and make sense to you

best regards

@ keira

Thanks for your help but can you solve this one for me too, its same as above only the language has been changed , its NONE now .

If it states that both are not blue then one can be blue , try this one below.


A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that NONE are blue?[/quote]

hello Joy
thanks for asking me . in fact the way the question is asked has double meaning and it is
a little ambigious
it is noone are blue it is what I have done :27/28


but if it is that both can not be blue
that is the probability to have noNE blue + the probablity to have maximum one blue and hence one of any other colors

hence or case

27/28+ 2/8x6/7

Hope it is right and make sense to you

best regards[/quote]


I quess the these two questions should clear the concept

1)Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that NONE of them are blue?

answer: (4/7) x (3/6) x (2/5) = 4/35


2)Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?


A. 5/7
B. 23/24
C. 6/7
D. 34/35
E. 8/13

so should we work like this : probability of all blue 3/7*2/6*1/6=1/35
none blue : 1-(1/35)= 34/35

for none you do not have to subtract from one

The question asks " they will both be not blue". Considering that these are draws without replacement there are following possibilities :
draw 1, draw 2
===== =====
1) Not Blue, Blue;
2) Blue, Not Blue;
3) Blue, Blue;
4) Not Blue, not Blue;

The answer has to be either P(4) OR all exclusion of the 3 previous conditions i.e. 1- [P(1) + P(2) + P(3)] => neither of the draws are blue

P(4) = 6/8*5/7=15/28

1- [P(1) + P(2) + P(3)] = 1 - [ (6/8 * 2/7) + (2/8*6/7) + (2/8*1/7)] = 1 - [26/8*7] = 1 - [13/28] = 15/28

both approaches have same answer

Probability that both are not blue means both are other colors(except blue)

6C2 / 8C2 = 15/28

Answer = 6C2 / 8C2
=15/28

Total ways of selecting card = 8C2
Favorable = 6C2 (neglecting 2 blue cards)

PS: You did not consider the case of atleast one blue. which can be picked 12 ways.
=1-2C2/8C2-12/8C2
=1-(1/28)-(12/28)
=(28-1-12)/28
=15/28

A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16

This is a post from one of the earlier posts in the forum. All the replies assumed that the cards are drawn one by one, but this assumption is not mentioned in the question.
Why cannot we draw 2 cards together and get the answer as 6/8.

Hi sd.1223, I also had the same problem

However, the default assumption is that cards are always drawn one by one unless there is clear indication that they will be drawn together / drawn with replacement.