Seeing The Red Pin

The number of ways to choose 10 pins from 15 = $\dbinom{15}{10}$ .

Number of ways to choose 1 red pin and 9 white ones = $\dbinom{14}{9} \times 1$

Probability of having a red pin in a ten pin arrangement = $\frac{\dbinom{14}{9}}{\dbinom{15}{10}} = \frac{2}{3}$

As there are 7 positions the red pin could be and 9! ways for the white pins and 10! ways to arrange ten pins, the probability of the red pin being visible is $\frac{ 7 \times 9!}{10!} = \frac{7}{10}$ .

Therefore, the probability of selecting a red pin and it being visible is $\frac{2}{3} \times \frac{7}{10} = \frac{7}{15}$ .

$7 + 15 = \boxed{22}$

There are fifteen total pins, and seven places where the red pin could be placed - the odds are therefore $7/15$ which causes $a+b=\boxed{22}$ .

First, we find the probability that the red pin is chosen from the $15$ available pins.

That is equal to $\dfrac{\dbinom{14}{9}}{\dbinom{15}{10}} = \dfrac{2}{3}$

Now, the probability of the bowler seeing the pins depends on if the pins are set in the $7$ favourable positions. There are exactly $3$ positions in the triangle where the bowler cannot see the pins. The probability of this happening is $\dfrac{7}{10}$

The probability of these events happening together is $\dfrac{2}{3} \times \dfrac{7}{10} = \dfrac{7}{15}$ giving us $7 + 15 = \boxed{22}$

Total available combinations= 15C10 ( choose 10 from 15) = 3003 To choose one red pin from 15 pins = total of all - (those with no red pins No red pin means ALL white pins = 14C10 ( or choose only 10 from 14 whites)= 1001 Therefore, to have one red pin from 15 pins is 3003-1001= 2002 The probability of choosing a red pin in the 10 pin combination is 2002/3003.

Therefore, the probability of seeing the red pin is 2002/3003 * 7/10= 14014/30030=7/15

There are 15 pins. After any given setup, 7 are visible. So the probability that the red pin is one of those is 7/15.

Total number of ways in which 15 pins can be arranged = 15!. Favourable outcomes are where one of the red pin is in the 7 visible slots, once the red pin is placed, we have 14! ways of arranging the remaining white pins. Therefore, Probability of red pin being visible = $\frac{7 \times 14!}{ 15!} = \frac{7}{15}$ .

Suppose bowling pins were arranged in triangular array of 15. Of these, the red pin must be in 7 places to be "seen". Thus the probability is 7/15.

Probability of choosing a red pin is 14C9/15C10 = 2/3 probability of seeing the pin is 7/10. so probability of seeing the red pin is 2/3 x 7/10 = 7/15 so a + b = 22

First, the probability that the collection contains the red pin is

$1-P($ all black $)=1- \frac{{15 \choose 14} }{ {15 \choose 10}} = \frac{2}{3}$ .

From a frontal view, three of the ten pins cannot be seen.

The probability that a red pin is visible is thus:

$\frac{2}{3} \cdot \frac{7}{10} = \frac{7}{15}$ ,

which gives the answer $7+15= \boxed{22}$ .

Probability of RED pin being in the triangle: $\frac{10}{15}$

Probability of being able to see the red pin, within the triangle: $\frac{7}{10}$

Multiplying the above two probabilities gives the probability of the pin setter choosing the red pin and placing that red pin where the bowler can view it.

$\frac{10}{15} \times \frac{7}{10} = \frac{7}{15}$

$7+15=22$

probability for the red pin to be in 10 pins chosen by pinsetter = 10/15 = 2/3
probability that pin is visible from 10 pins chosen = 7/10 (as 7 pins are visible)

Total probability = 2/3*7/10 = 7/15 a=7 and b = 15 ; a+b = 22

Imagine a line pins, $14$ white and $1$ red. We say the red pin "shows" if it is in one of the first $7$ spots. Thus, the desired probability is $\frac{7}{15}$ .