GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
In college basketball, a player awarded a one-and-one gets to shoot
[#permalink]
30 May 2022, 03:38
30 May 2022, 03:38
Expert Reply
4
Bookmarks
00:00
Question Stats:
22% (02:22) correct
77% (02:10) wrong based on 9 sessions
Hide Show timer Statistics
In college basketball, a player awarded a “one-and-one” gets to shoot one free throw and, if successful, shoot a second one. Each successful free throw is worth one point. Suppose a player succeeds at free throws 80% of the time. On average, how many points should this player expect to score for each one-and-one?
1
1.16
1.28
1.44
1.6
1
1.16
1.28
1.44
1.6
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
Re: In college basketball, a player awarded a one-and-one gets to shoot
[#permalink]
23 Jun 2022, 14:00
23 Jun 2022, 14:00
1
Expert Reply
1
Bookmarks
OE
Quote:
The player might earn one point by making the first shot (80% chance) and failing the second (20% chance). The likelihood of this is 0.8 × 0.2 = 0.16. The player might also earn two points by making both shots, with a likelihood of 0.8 × 0.8 = 0.64. The average expected number of points per one-and-one awarded is thus 0.16 × 1 + 0.64 × 2 = 1.44.
Re: In college basketball, a player awarded a one-and-one gets to shoot
[#permalink]
27 Jul 2022, 19:18
27 Jul 2022, 19:18
Carcass wrote:
OE
Quote:
The player might earn one point by making the first shot (80% chance) and failing the second (20% chance). The likelihood of this is 0.8 × 0.2 = 0.16. The player might also earn two points by making both shots, with a likelihood of 0.8 × 0.8 = 0.64. The average expected number of points per one-and-one awarded is thus 0.16 × 1 + 0.64 × 2 = 1.44.
Thank you
