Jtdylktuy

Forgive me for anything that I Write wrong since this is still new to me and I haven’t used stackexchange in a long time...
Also I’m writing this using my ipad ....

A psychometric exam is an Israeli exam that was based on the SAT (but you are not allowed to use a calculator during the exam)

What is the fastest way to solve this question (well I don’t really know Anyway to solve the question)

The question says :

A bag contains 3 blue , 3 white , 3 red balls
What is the probabilty to take 3 balls out of the bag (without returning the balls back to the bag) randomly and each ball should have a different color ?

I tried doing this :

3/9 * 3/8 * 3/7 => 3/56

The suggested answers are

1) 3/28

2) 5/28

3) 15/28

4) 9/28

What did I do wrong trying to solve the question ?

You should review basic Combinatorics: Permutations and Combinations. The calculations are fairly simple, and the topics/applications come up very frequently.

The idea is to count up the total number of ways you can make a choice, and then the total number of ways you can make a choice that fits the conditions you want.

Taking 3 objects from a group of 9 without respect to order is a Combination.

The notation is: $9choose 3$$=frac{9!}{3! 6!} =84$

So there are $84$ ways to choose $3$ balls from the $9$.

Now, you want them all to be of different colors.

So count the ways this can happen: 3 ways one ball can be blue, times 3 ways one ball can be red, times 3 ways one ball can be white.

So there are $27$ ways to get the desired RWB configuration.

Your probability is then: $frac{27}{84} = frac{9}{28}$

What you have done makes sense.
Let us first select the red ball: $frac{3}{9}$ ways of doing this. Then the white ball $frac{3}{8}$ ways of doing this. Then the blue ball $frac{3}{7}$ ways of doing this.

So we can conclude that there are $frac{3}{56}$ ways of picking the red ball, then a white ball and then the blue ball. But why did we pick red, white blue? We could have done white, blue, red. Or blue, red, white. So how many different ways can we do this?

6 different ways. RWB, RBW, WBR, WRB, BWR, BRW.

Each one of these ways has a probability of $frac{3}{56}$.

So the final answer should be

$6times frac{3}{56}$.

For the balls to all be different colors, each needs to be a different color from the balls already chosen. So we can multiply the probabilities at each stage that a ball of a different color from all previous ones is drawn.

• Ball 1: You haven't chosen any balls yet, so any ball works. Probability 1.


• Ball 2: There are 6 balls that don't match the color of the first ball and 8 balls total. Probability 3/4.


• Ball 3: There are 3 balls left that don't match either of the first two colors and 7 balls total. Probability 3/7.

$1 times 3/4 times 3/7 = 9/28$