Probability please?

Question

a basketball player hits his shot 44% of the time. If he takes four shots during a game, what is the probability that he misses the first and makes the next three shots?

Please tell me how you guys can make sense to calculate this..

Answer 1

Since the outcome of one shots has no effect on future shots, you need to think of each of the 4 shots as a separate event. Then, to find the probability of all 4 events happening together, we can multiply the probabilities of each of the individual events.

Shot 1 - P(Miss) = 56% chance (because he makes 44%, so 100 - 44)
Shot 2 - P(Make) = 44% chance
Shot 3 - P(Make) = 44% chance
Shot 4 - P(Make) = 44% chance

P(Miss, Make, Make, Make) = .56 * .44 * .44 * .44 = 0.0477 or 4.77%

So he has just under a 5% chance of missing his first shot but making the following three.

Answer 2

i've got faith my argument will not be properly gained. I say the possibility is 50% permit a ??, permit b ?? and randomly opt for the values for a and b. As already pointed out, for a ? 0, P( a < b²) = a million, it truly is trivial. purely slightly much less trivial is the thought that P(a < 0 ) = a million/2 and in this occasion P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what happens whilst a > 0 For a > 0, whilst it extremely is easy to coach there's a non 0 possibility for a finite b, the decrease, the possibility is 0. a < b² is corresponding to announcing 0 < a < b², keep in mind we are purely finding at a > 0. If this a finite era on a limiteless line. The possibility that a is a factor of this era is 0. P( a < b² | a > 0) = 0 As such we've a finished possibility P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 keep in mind, it truly is because of the endless instruments. no count what kind of era you draw on paper or on a working laptop or computing gadget you will hit upon a finite possibility that looks to mindset a million. yet it truly is as a results of the finite random huge style turbines on the computing gadget and if we had this query requested with finite values there may well be a a answer greater desirable than 50%. i don't advise to be condescending, yet please clarify why employing the Gaussian to approximate a uniform distribution is a very good thought? are not endless numbers relaxing. Cantor whilst mad working with them! :)

Answer 3

I am also not good at this but i will make a try:

The probability that a shot is a
HIT = 1/2
MISS = 1/2

The basketball player HITS his shot 44% of the time, consequently MISSES 56% of the time

The probability that he MISSES the first shot is

1/2 x 56% = 28%
The probability that he HITS the first shot is
1/2 x 44% = 22%

The probability that he misses the first and hits the 3 shots is
(28%)(22%)(22%)(22%) =
(.28)(.22)(.22)(.22) = 0.002981 x100 = .2981%

Answer 4

44% makes the shot, that means 56% miss

the probability of getting the first shot missed is 56%
the probability of making the 3 shots is 44%

.56 * (.44)^3 = .0477 or 4.77%

Answer 5

If events are independant...you can multiply the probabilites together.

44% chance to make a shot = 56% chance of missing.

So miss, make, make, make
=
miss*make*make*make
(suppose to be P(miss) and P(make))

.56 * .44 * .44 * .44= whatever everyoen else wrote as the number.

Answer 6

P (hit)=44%=.44
therefore P (miss)=1-P(hit)=1-.44=.56.
Now,
P(miss,hit,hit,hit,hit)=p(miss)*P(hit)*P(hit)*P(hit)*
=.56*.44*.44*.44
=0.0477
=4.77%

Answer 7

P = (.56)(.44)^3 = 0.0477

Answer 8

Each shot has its own discrete probability, then you multiply them together
miss x hit x hit x hit
(1-0.44) x (0.44) x (0.44) x (0.44)
= 0.048
4.8%

Answer 9

P(hits a shot) = 0.44

We want:

P(Miss,hit,hit,hit) = (1-0.44) * 0.44 * 0.44 * 0.44

= 0.04770304