some distribution/probability problems

[mcrae]

1) A player scores on 85% of foul shots. If he takes 8 foul shots in a game, what is the probability that he scores on exactly 7 of them?

My answer: 0.85^7

2) One out of 6 blocks is blue. A customer wants to buy two blue blocks. What is the probability of finding exacly two such blocks among the first four tested?

My answer: (4C2)(5/6)(5/6)(1/6)(1/6)=.114, but i think I messed up due to the wording of the problem.

3) Forty percent of the chocolates are red. The production line mixes chocolates randomly and packages 10 per box. Find:

a) the probability that less than four candies in a given box are red.

My answer: This means that 0, 1, 2, or 3 are red, so I added those up:

0.6^10 + (10C1)(0.6^9)(0.4^2) + (10C2)(0.6^8)(0.4^2) + 10C3(0.6^7)(0.4^3) = 0.38

b) the probability that exactly four candies in a given box are red.

My answer: (10C4)(0.4^6)(0.6^4) = 0.252

4) A jet engine has 0.0001 probability of failure while in flight. For a jet with four engines, what is the probability of at least two of them failing?

My answer: 1 - p(0) - p(1)

1 - 0.9999^4 - 4(0.0001)(0.9999^3) = 5.9992 x 10^-8

5) Use the expansion of (a+b)^3 to show how the binomial theorum is related to the binomial probability distribution.

My answer: I'm not too sure what the question wants. I've fiddled around for an hour and come up with this:

term x of the expression (a + b)^n is equal to the probability of n-x given n trials and probability of success a and probability of failure b

ex: (a + b)^3, a = 0.85, b = 0.15
term 0:
(3C3)(a^(3 - 3))(b^3)
3(0.85^2)(0.15)

probability of 3:
(3C3)(0.85^3)(0.15^3-3)

 

[Denis]

1) A player scores on 85% of foul shots. If he takes 8 foul shots in a game, what is the probability that he scores on exactly 7 of them?
My answer: 0.85^7

.85^7 = .3207...
What are you doing?

 

[soroban]

Hello mcrae!

Good work . . . a few errors . . .

1) A player scores on 85% of foul shots.
If he takes 8 foul shots in a game, what is the probability that he scores on exactly 7 of them?

My answer: \(\displaystyle 0.85^7\;\) . . . no

"Exactly seven scores" means: seven scores and one miss.

\(\displaystyle P(\text{exactly 7 scores})\:=\_8C_7)(0.85)^7(0.15) \:\approx\:0.3847\)

2) One out of 6 blocks is blue. A customer wants to buy two blue blocks.
What is the probability of finding exacly two such blocks among the first four tested?

My answer: \(\displaystyle (_4C_2)\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)^2\;=\;0.114\) ?

Your set-up is correct . . . but the answer is: \(\displaystyle \,0.115748741\)

3) Forty percent of the chocolates are red.
The production line mixes chocolates randomly and packages 10 per box.
Find:

a) the probability that less than four candies in a given box are red.

My answer: This means that 0, 1, 2, or 3 are red, . . . Right!
so I added those up:

\(\displaystyle 0.6^10\,+\,(_{10}C_1)(0.6)^9(0.4)^2\,+\,(_{10}C_2)(0.6)^8(0.4)^2\,+\,(_{10}C_3)(0.6)^7(0.4)^3 \;= \; 0.38\;\) . . . Yes!

b) the probability that exactly four candies in a given box are red.

My answer: \(\displaystyle \_{10}C_4)(0.4)^6(0.6)^4\;=\;0.252\;\) . . . no

You found the probability that six are red.

\(\displaystyle P(\text{4 red, 6 others})\;=\;(_{10}C_4)(0.4)^4(0.6)^6\;\approx\;0.2508\)

4) A jet engine has 0.0001 probability of failure while in flight.
For a jet with four engines, what is the probability of at least two of them failing?

My answer: \(\displaystyle \,1\,-\,P(0)\,-\,P(1)\;\) . . . yes!

\(\displaystyle 1\,-\,(0.9999)^4\,-\,4(0.0001)(0.9999)^3\:=\:5.9992\,\times\,10^{-8}\;\) . . . Right!

5) Use the expansion of \(\displaystyle (a\,+\,b)^3\) to show how the binomial theorum
is related to the binomial probability distribution.

Your interpretation is correct.

I would explain it like this . . . (I don't claim that it's better, though.)

Let \(\displaystyle a\) = probability of success \(\displaystyle (S)\)
Let \(\displaystyle b\) = probabiity of failure \(\displaystyle (F)\)

The expansion of \(\displaystyle (a\,+\,b)^3\,\) gives the probabilities of all possible outcomes.

\(\displaystyle (a\,+\,b)^3\;=\;\;a^3\;\;+\;\;\;3a^2b\;\;\;\;+\;\;\;3ab^2\;\;\;+\;\;b^3\)
. . . . . . . . . \(\displaystyle \overbrace{P(3S)}\;\;\overbrace{P(2S,1F)}\;\;\overbrace{P(1S,2F)}\;\;\overbrace{P(3F)}\)

 

[mcrae]

thanks soroban. the explanation for the last one is great. i seem to always mess up on questions like similar to the foul shot one lol.

abotu the last question, for anyone wondering, my i believe my examples were wrogn even if my statement was right.