GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Oct 2019, 13:52

### GMAT 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# On every game, there are only two cases: the player wins or loses

Author Message
TAGS:

### Hide Tags

Retired Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1331
Location: Viet Nam
On every game, there are only two cases: the player wins or loses  [#permalink]

### Show Tags

11 Dec 2016, 05:50
2
1
00:00

Difficulty:

95% (hard)

Question Stats:

43% (02:55) correct 57% (02:29) wrong based on 86 sessions

### HideShow timer Statistics

On every game, there are only two cases: the player wins or loses. The probability that Michael wins on each game is 0.4. How many games does Michael have to play at least so that the probability that Michael wins on at least one game is greater than 0.95?

A. 4
B. 5
C. 6
D. 7
E. 8

_________________
Senior Manager
Joined: 13 Oct 2016
Posts: 359
GPA: 3.98
Re: On every game, there are only two cases: the player wins or loses  [#permalink]

### Show Tags

11 Dec 2016, 09:22
2
Hi

P(winning at least one game)>0,95 is the same as P(loosing all)<0,05

$$0,6^n < 0,05$$

$$n=5 ----> 0,0777...$$

$$n=6 ----> 0,04665...$$

Senior Manager
Joined: 13 Oct 2016
Posts: 359
GPA: 3.98
Re: On every game, there are only two cases: the player wins or loses  [#permalink]

### Show Tags

11 Dec 2016, 23:07
Another way to calculate:

$$0.6^n < 0.05$$

$$ln (6^n) < ln(0.05)$$

$$n*ln(0.6) < ln(0.05)$$

$$n*(-0.5108) < -2.9957$$

$$n > 5.864..$$

Min integer value of $$n$$ is $$6$$.

Although it's hard to be done without calculator
Intern
Joined: 12 Dec 2016
Posts: 1
Re: On every game, there are only two cases: the player wins or loses  [#permalink]

### Show Tags

13 Dec 2016, 13:27
Probability of wining AT LEAST one game is 1 - probability of wining no games:

1 - P(0) = P(>=1)

If we want the probability of wining at least 1 games to be at least 95% (>=0.95); then probability of wining no games has to be less than or equal to 5% (<=.05).

Probability of losing a game [P(0)] is (6/10)^ n, where n is the number of games played.

We start with option c to know if we should go up or down if c is false.

(6/10)^6 = 45,792/1000000= 0.045792 chance that you lose 6 games in a row

The probability of winning at least 1 game is 1 - P(0)
1 - 0.045792 > .95

Retired Moderator
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 483
Re: On every game, there are only two cases: the player wins or loses  [#permalink]

### Show Tags

18 Aug 2017, 07:16
A similar tough and tricky question is here

https://gmatclub.com/forum/tough-and-tr ... 85179.html
_________________
Hasan Mahmud
Non-Human User
Joined: 09 Sep 2013
Posts: 13421
Re: On every game, there are only two cases: the player wins or loses  [#permalink]

### Show Tags

10 Mar 2019, 09:59
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: On every game, there are only two cases: the player wins or loses   [#permalink] 10 Mar 2019, 09:59
Display posts from previous: Sort by