Probability Intro

It's been a long time and I'm in an Intro to Probability and Statistics class as a requirement. Even referring to my notes from last class is not helping. Here is the problem.
A particular basketball player hits 70% of her free throws. When she tosses a pair of free throws, the 4 possible simple events and 3 of their associated probabilities are as given below:

Simple Event Outcome of 1st Free Throw Outcome of 2nd Free Throw Probability
1 Hit Hit .49
2 Hit Miss ?
3 Miss Hit .21
4 Miss Miss .09

a. Find the probability that the player will hit on the 1st throw and miss on the 2nd throw. Figuring it out algebraically, I arrived at .29, or 29%. What I am unsure of is how to write it out in probability language.

b. Find the probability that the player will hit on at least one of the two free throws. I'm totally lost on this one.

Thanks so much,
Sandra

I will try to help

there are 4 possible outcomes, and their probabilities must add to 1 . Why 1. because these are all possibilities.
in the notation you used

HH has a probability of P[hh]=.49
HM has a probability of P[HM]=x we will figure it out in a minute\
MH has a probability of P[MH]=.21
MM has a probability of P[MM]=.09
but the sum must add to 1
1=PHH]+P[HM]+P[MH]+P[MM] substitute values
1=.49+x+.21+.09
1=.79+x
.21=x answer

probability of .21 that the shooter miss the first shot but make the second one.
========================================================================================
What is the probability the shooter will hit the 1st shot And miss the second,[MH in your notattion] OR miss the first and hit the seccond [HM in your notation.
Stated simpler
What is the probability of MH OR HM
When you say OR you add. When you say AND you multiply. We said MH OR HM so we add
P[HM]+P[MH]= .21+.21
P[MH]or P[HM]=.42 answer

Hope the wordage helped

Arthur

1 Hit Hit .49
2 Hit Miss ?
3 Miss Hit .21
4 Miss Miss .09

a. Find the probability that the player will hit on the 1st throw and miss on the 2nd throw. Figuring it out algebraically, I arrived at .29, or 29%. What I am unsure of is how to write it out in probability language.

b. Find the probability that the player will hit on at least one of the two free throws.

Click to expand...

Arthur gave you the answer for problem a): Hit, Miss is .21

In part b, the problem reads "at least one", meaning HH, HM, and MH are all to be counted. The probability is the sum of these three:

.49 + .21 + .21 = .91

arthur ohlsten said:

I will try to help

there are 4 possible outcomes, and their probabilities must add to 1 . Why 1. because these are all possibilities.
in the notation you used

HH has a probability of P[hh]=.49
HM has a probability of P[HM]=x we will figure it out in a minute\
MH has a probability of P[MH]=.21
MM has a probability of P[MM]=.09
but the sum must add to 1
1=PHH]+P[HM]+P[MH]+P[MM] substitute values
1=.49+x+.21+.09
1=.79+x
.21=x answer

probability of .21 that the shooter miss the first shot but make the second one.
========================================================================================
What is the probability the shooter will hit the 1st shot And miss the second,[MH in your notattion] OR miss the first and hit the seccond [HM in your notation.
Stated simpler
What is the probability of MH OR HM
When you say OR you add. When you say AND you multiply. We said MH OR HM so we add
P[HM]+P[MH]= .21+.21
P[MH]or P[HM]=.42 answer

Hope the wordage helped

Arthur

Click to expand...

Don't you think we missed the given fact that 'player hits 70% of her free throws'?
I am not sure this 0.7 refers to success of both first and second throw (which means HH + HM = .49 + .21 = .7)
Or it refers to the each throw (which means single throw .7 but HH = .7 x 2 = .49)
We can agree that both first throw and second throw are mutually exclusive (they are independent and have nothing to do with each other, first one result has no effect on the other), no?
P(first throw ∪ second throw) = P(first throw) + P(second throw) - P(first throw ∩ second throw) = P(first throw) + P(second throw) - 0 = P(first throw) + P(second throw)
Anybody can clarify this?

Mojna is responding to an old post (I'm just pointing this out for the benefit of the other helpers). Hopefully it can help with Mojna's understanding of probability...

Mojna said:

I am not sure this 0.7 refers to success of both first and second throw (which means HH + HM = .49 + .21 = .7)
Or it refers to the each throw

Click to expand...

The 0.7 refers to each individual throw.
P(both first and second throw hit) = P(HH) = 0.7 * 0.7 = 0.49

Mojna said:

We can agree that both first throw and second throw are mutually exclusive (they are independent and have nothing to do with each other, first one result has no effect on the other), no?

Click to expand...

Agreed. This just means that she isn't affected by the success or failure of her first shot. The chance of success on her second shot is still 70% regardless.

The probability space for the op is two shots. You must agree that she could hit both the first and second throws? Therefore, the parts highlighted red below are incorrect...

Mojna said:

P(first throw ∪ second throw) = P(first throw) + P(second throw) - P(first throw ∩ second throw) = P(first throw) + P(second throw) - 0 = P(first throw) + P(second throw)

Click to expand...

P(first throw ∪ second throw) = P(hitting one or more of the two shots)
P(first throw ∩ second throw) = P(hitting both shots)

Hope this helps. I recommend that you draw a Venn diagram if you're still confused.

Mojna said:

Don't you think we missed the given fact that 'player hits 70% of her free throws'?
I am not sure this 0.7 refers to success of both first and second throw (which means HH + HM = .49 + .21 = .7)
Or it refers to the each throw (which means single throw .7 but HH = .7 x 2 = .49)
We can agree that both first throw and second throw are mutually exclusive (they are independent and have nothing to do with each other, first one result has no effect on the other), no?
P(first throw ∪ second throw) = P(first throw) + P(second throw) - P(first throw ∩ second throw) = P(first throw) + P(second throw) - 0 = P(first throw) + P(second throw)
Anybody can clarify this?

Click to expand...

The 0.7 is the probability of a hit on any one throw, and throws are assumed to be independent. We can tell that from the numbers provided in the table.

There are many ways to solve this, because too much information was given. It seems likely that the OP was not expected to know much yet, and just to make the sum be 1 for the first part (1-(0.49+0.21+0.09) = 0.21), and then to add up the relevant individual probabilities for the second (0.49+0.21+0.21 = 0.91).

But you're right that, without being given the table at all, would could find P(HM) = P(H)*P(M) = 0.7*0.3 = 0.21 for the first page; and then, for the second part, either (a) just find the three individual probabilities the same way and add them, or (b) use the formula for the union; or (c) use 1 - P(MM) = 1 - 0.3^2 = 0.91.