Draw The Ball

Probability of getting atleast 1 red ball (1,2 or 3 red balls) is the same as 1-(probability of getting no red ball)

Probability of getting no red ball: (Using combinations)

n(Event) = $\frac{13!}{3!10!}$ = 286

n(Sample Space) = $\frac{16!}{13!3!}$ = 560

Probability = $\frac{n(E)}{n(S)}$ = $\frac{286}{560}$

Probability of getting at least one read ball = 1 - (probability of getting no red ball)

Required probability = $1 - \frac{286}{560}$ = $\frac{137}{280}$

a=137 and b=280

a + b = 417

The easiest way to solve this problem is to begin with the situation we wouldn't want, in this case avoiding red completely. Since we're starting with 16 balls, our odds of avoiding red on the first pick are 13/16. Then, should we have avoided red, these change to 12/15 and 11/14 for the second and third chances respectively. To find the chance overall, we must multiply the fractions together, giving 143/280. Now, if that fraction represents the chances of drawing no red balls, we need to do 280-143 to find the chances of 1 or more, which is 137(/280). 137+280 is 417, which is our answer.

There are a total of $\binom{8+5+3}{3}$ ways of drawing 3 balls without regard to order. The number of ways of $not$ drawing a red ball is given by $\binom{8+5}{3}$ since we are choosing 3 balls from only the yellow and white balls. Therefore the probability of drawing at least one red ball is $1-\frac{\binom{13}{3}}{\binom{16}{3}}=1-\frac{286}{560}=\frac{137}{280}=\frac{a}{b}$

$a+b=\boxed{417}$

The probability that we do not draw any red balls is equal to $\frac{13}{16} \times \frac{12}{15} \times \frac{11}{14} = \frac{143}{280}$ .So the probability that we draw at least 1 red ball is equal to $1 - \frac{143}{280} = \frac{137}{280}$ .So $a = 137,b = 280$ and $a + b =417$ .

This is hypergeometric distribution question.

Let X be a number of red picked up.

Using axiom of probability, P(X=1)+P(X=2)+P(X=3) = 1-P(X=0)

P(X=0)=1716/3360

Therefore, 1-P(X=0)= 1644/3360.

The GCD is 12, then 1644/3360 = 137/280.

137+280 = 417

$P(\text{atleast one red ball}) = 1 - P(\text{no red ball})$

$P(\text{no red ball}) = \dfrac{13}{16} \times \dfrac{12}{15} \times \dfrac{11}{14}$

$P(\text{atleast one red ball}) = 1- \left( \dfrac{143}{280} \right) = \dfrac{137}{280} = \boxed{417}$

I solved it the other way around , here's what I did :

First of all , the sample space = 16C3 = 560

Possibilities of only 1 red ball drawn = ( 13C2 ) * ( 3C1 ) = 234

Possibilities of 2 red balls drawn = ( 13C1 ) * ( 3C2 ) = 39

Possibiliteis of 3 red balls drawn obviously = 1

Therefore the possibilities of at least one red ball drawn = 234 + 39 + 1 = 274

Then divide it by the sample space : 274/560 = 137/280 = a/b Then : a + b = 137 + 280 = 417 ;) :D

total no. of ways of taking 3 balls from 16 balls= 16c3 = 560

no. of ways when no red ball is taken= 13c3 = 286

so, no. of ways when atleast one red ball is taken = 560-286= 274

probablity that we draw atleast a red ball= 274/560=137/280.

a= 137 & b=280.

theirfore, a+b= 280+137 =417

number of ways of selecting 3 balls out of 16 = 16C3 = 560

number of ways of selecting 3 non-red balls = 13C3 = 286

probability of at least one red ball = (560 - 286)/560 = 137/280

137 + 280 = 417

Out of a total of 16 balls, let us consider the cases in which we do not get any red ball at all. For the first ball, we have a total of $8+5 = 13$ choices out of $16$ (since we are excluding the red balls); for the second, we have a remaining total of $12$ choices out of $15$ ; for the third ball, we have $11$ choices out of $14$ .

$P(No Red Balls) = \frac{13}{16} \times \frac{12}{15} \times \frac{11}{14} = \frac{143}{280}$

And so, $P(At Least One Red) = 1 - \frac{143}{280} = \frac{137}{280}$

We have $a = 137$ and $b = 280$ , giving us $a + b = 417$ , which is the required answer.

There are two ways...you can make different cases or simplest way is find the probability of getting only yellow and white balls and subtract it from 1(complement).....

Because counting the number of ways to count AT LEAST 3 red balls is too hard we are going to use complementary counting to count how many ways we can not count a red ball. The answer we want to get is now $1-\frac{{13 \choose 3}}{{16 \choose 3}}$ because the amount of success in this case is choosing the amount of balls that are not red over the amount of total ways you can choose a ball.

Let's imagine we only have the $8$ yellow balls and the $5$ white balls. So the combinations are $\frac {13!}{3!\cdot 10!} = 286$ . Now let's add up the $3$ red balls. The combinations are $\frac {16!}{3!\cdot 13!} = 560$ . That means that the difference between this two cases, $560 - 286 = 274$ , are the combinations that involves at least $1$ red ball. But remember that we have $16$ balls, so the total of possible combinations are $560$ . The answer is $\frac {274}{560} = \frac {137}{280}$ , and so $a + b = 137 + 280 = \boxed {417}$ .