GMAT Math

e12star wrote:Question from Word Translations (Manhattan GMAT), chpt 5:
A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts two flowers together at random in a bouquet. However, the customer calls and says that she does not want two of the same flower. What is the probability that the florist does not have to change the bouquet?

ronnie1985 wrote:I think there is a serious flaw in the reasoning as now the bouquet is already having 2 flowers of the same color. The florist has to take one of the flower, the color of which is not known and replace it with another flower of some other color. This calls for Baye's Theorem.

Cumulonimbus wrote:Hi Brent,

I tried it as below:

P(A and B) = (2/9)*(3/8) = 6/72
P(B and P) = (3/9)*(4/8) = 12/72
P(P and A) = (4/9)*(2/8) = 8/72

Probability of choosing different flowers = P(A and B) + P(B and P) + P(P and A)
= (6+12+8)/72 = 26/72 = 13/36.
Why is this method not giving correct answer?
How is this different from Baye's theorem?

Also, these flower types can come in any order - AB, BP, PA, BA, PB, AP - so total 6 cases,
Are all these cases different, is AB really different from BA?

Since correct answer is 13/18, do i need to multiply here by 2 because AB and BA are different?

My confusion arises in deciding when order matters and when it does not.

Another similar question:

"A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?"

Can you please help with this. Are these two questions any different?

KR,

Cumulonimbus wrote:Hi Brent,

Thanks.
However I still have one point unclear.
How to decide when order matters and it does not.

For this case - as per my understanding - either first choosing an A and then B or first choosing a B and then A, ultimately results in one type of arrangement.

Cumulonimbus wrote:
Below examples are from Manhattan guide.
Example 1:
An office manager must choose five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition is allowed. How many different lock codes are possible?
Manhattan guide way - number of decisions to be made:
5*8*7*6*4=6720.

My way: I tried via Combinatorics,

Number of ways in which 2 odd digits can be chosen from 5 = 5C2 = 10 (this assumes all five are identical, the two chosen ones as well as the three not-chosen ones).
However since the two digits chosen are not identical, I need to multiply it by 2!, number of ways the chosen two digits can be arranged, so
Number of ways in which 2 odd digits can be chosen from 5 = 5C2 = 10*2 = 20

Similarly, for the rest 3 digits, number of ways to choose 3 digits out of 8 = 8C3, and here I need to multiply it by 3! to give = 56*3! = 336,

Thus total no of codes/combination = 20*336 = 6670.

Here order matters.

Example 2:
The I Eta Pi fraternity must choose a delegation of three senior members and two junior members for an annual fraternity conference. If I Eta Pi has 6 senior members and 5 junior members, how many different delegations are possible?

Here answer is simply = 6C3 = 20, because order does not matter here.

Please guide.

KR,

Deepthi Subbu wrote:My soln :

2/9*3/8 + 2/9*4/8 + 3/9*4/8 => 13/36 . However the answer is 13/18. Where am I going wrong?
Since order does not matter here, as azalea and then followed by buttercup is the same otherwise.And in addition to this if there are 3 ways to choose the placement, Il be in more trouble.