## Amazon Interview Question

Software Engineer / Developers**Country:**United States

**Interview Type:**Written Test

You forgot a factor of 2 in the 8!/3*5! term it should instead be 8!/(3!*5!). Your numerical answer is correct though.

Can someone tell me why is there a '1' in the numerator (3!/2! + 1)/2^3.

I thought is should just (3!/2!)/2^3

I would say the second challenge.

The first challenge requires you to make it 66% of the time while the second challenge only requires you to make it 63%. The drop of 3% might be significant.

Also, the condition for independence between shoot is questionable. As you try 8 times, you have more practices and might be able to make it into the basket in the subsequent throws. Therefore, the option of trying 8 times might mean better chance of winning the game.

That's what I think from my statistics perspective.

Amount of ways to win challenge 1:

`C(3, 2) + C(3, 3)`

, amount of ways to lose challenge 1:

`C(3, 0) + C(3, 1)`

. But

`C(n, k) === C(n, n-k)`

, so

`C(3, 2) === C(3, 1), C(3, 0) === C(3, 3)`

.

Amount of ways to win challenge 2:

`C(8, 5) + C(8, 6) + C(8, 7) + C(8, 8)`

and amount of ways to lose challenge 2:

`C(8, 0) + C(8, 1) + C(8, 2) + C(8, 3) + C(8, 4)`

, but

`C(8, 8) === C(8, 0), C(8, 1) === C(8, 7), C(8, 2) === C(8, 6), C(8, 3) === C(8, 5)`

so, amount of ways to lose in the second case are greater.

It depends on the probability of winning.

This is a classic example of bernoulli trial.

P(2 out 3) = (3 Choose 2) p^2 * (1-p)

P(5 out 8) = (5 Choose 8) p^5 * (1-p)^3

solving this:

P^3 * (1-P)^2 >= 15/56

Therefore, the second option is a better choice for the values of p that satisfy above condition.

First, compute probability of getting success (2 out of attempts):

1. Compute sample space i.e., All possible outcomes of 3 attempts: 8 BBB, BBN, BNB,...etc. Where B->Basket and N->No Basket.

2. Compute How many of the 8 outcomes above have atleast 2 Baskets in them.

Ans: 4 or 3C2+1 (1 for combination with all 3 baskets).

3. Now final step for challenge 1: Probablity of achieving success (2 out of 3 attemts)=4/8 or 0.5.

Now Challenge 2:

1. 8 attempts, possible outcome combinations are: 256.

2. Number of possible combinations of 5 baskets in these 256 outcomes: 8C5+8C6+8C7+8C8=93.

3.Now compute probability of success for challenge 2 i.e., 5 out of 8 attempts: 93/256= ~0.30

Hence, it is better to take Challenge 1 as Prob of success for it (0.5) is greater than the second challenge

It depends.

Let the probability to make a shot be p.

The probability to win challenge one is:

`(3 choose 2) p^2 (1-p) + (3 choose 3) p^3`

since you can win by either hitting two shots or all three shots.

For challenge 2 it is

`(8 choose 5) p^5 (1-p)^3 + (8 choose 6) p^6 (1-p)^2 +(8 choose 7) p^7 (1-p) + p^8`

since you can either make 8,7,6 or 5 of the shots to win.

If the probability p is below ~65% challenge one is the one you should choose, if it's above challenge two. So three point shots --> 1 , layups or free throws (if you are a good free throw shooter) --> 2

okay ... probability of winning first game 2/3

probability of winning second game is 5/8

i was taught in 8th grade that to compare two fractions you make their denominators same....

(so take LCM) and the probability of winning first game becomes : 16/24

and the second game becomes : 15/24

since 16/24> 15/24 so the person should play the first game.

okay ... probability of winning first game 2/3

probability of winning second game is 5/8

i was taught in 8th grade that to compare two fractions you make their denominators same....

(so take LCM) and the probability of winning first game becomes : 16/24

and the second game becomes : 15/24

since 16/24> 15/24 so the person should play the first game.

okay ... probability of winning first game 2/3

probability of winning second game is 5/8

i was taught in 8th grade that to compare two fractions you make their denominators same....

(so take LCM) and the probability of winning first game becomes : 16/24

and the second game becomes : 15/24

since 16/24> 15/24 so the person should play the first game.

Let,

A = the event that you make 2 out of 3 baskets

B = the event that you make 5 out of 8 baskets

p = the probability of a single success (i.e. you make a basket)

q = the probability of a single failure (i.e. you miss the basket)

Then you should always choose event A (the first option) over event B (the second option).

Note that:

3 choose 2 = 3!/[(3-2)!*2!] = 3

8 choose 5 - 8!/[(8-5)!*5!] = 56

Pr(A) = 3*(p^2)*(q^1)

Pr(B) = 56*(p^5)*(q^3)

And hence we should the first option over the second option when: Pr(A)/Pr(B) > 1

or equivalently when: Pr(B)/Pr(A) < 1

Note that we are indifferent between option 1 and option 2 when Pr(A)/Pr(B) = 1

Finally, note that:

```
Pr(B)/Pr(A) = [56*p^5*q^3] / [3*p^2*q]
= 56/3*p^3^q^2
<= 56/3*(1/2)^3*(1/2)^2
= 56/3*(1/2)^5
= 56/3*(1/32)
= 56/96
< 1
```

And therefore, since Pr(B)/Pr(A) < 1 for any value of p between 0 and 1, we conclude that the first option is always "better" (i.e. more likely) than the second option.

Let,

A = the event that you make 2 out of 3 baskets

B = the event that you make 5 out of 8 baskets

p = the probability of a single success (i.e. you make a basket)

q = the probability of a single failure (i.e. you miss the basket)

Then you should always choose event A (the first option) over event B (the second option).

Note that:

3 choose 2 = 3!/[(3-2)!*2!] = 3

8 choose 5 - 8!/[(8-5)!*5!] = 56

Pr(A) = 3*(p^2)*(q^1)

Pr(B) = 56*(p^5)*(q^3)

And hence we should the first option over the second option when: Pr(A)/Pr(B) > 1

or equivalently when: Pr(B)/Pr(A) < 1

Note that we are indifferent between option 1 and option 2 when Pr(A)/Pr(B) = 1

Finally, note that:

```
Pr(B)/Pr(A) = [56*p^5*q^3] / [3*p^2*q]
= 56/3*p^3^q^2
<= 56/3*(1/2)^3*(1/2)^2
= 56/3*(1/2)^5
= 56/3*(1/32)
= 56/96
< 1
```

And therefore, since Pr(B)/Pr(A) < 1 for any value of p between 0 and 1, we conclude that the first option is always "better" (i.e. more likely) than the second option.

I can win the first challenge with 50% probability:

And the second one with 36% probability:

- Daniel November 15, 2014