It is currently 19 Oct 2017, 11:59

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

VP
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1182

Kudos [?]: 885 [0], given: 54

Location: Viet Nam
On every game, there are only two cases: the player wins or loses [#permalink]

### Show Tags

11 Dec 2016, 05:50
3
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

41% (01:55) correct 59% (01:44) wrong based on 98 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
[Reveal] Spoiler: OA

_________________

Kudos [?]: 885 [0], given: 54

Senior Manager
Joined: 13 Oct 2016
Posts: 367

Kudos [?]: 395 [0], given: 40

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
3
This post was
BOOKMARKED
Hi

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

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

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

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

Kudos [?]: 395 [0], given: 40

Senior Manager
Joined: 13 Oct 2016
Posts: 367

Kudos [?]: 395 [0], given: 40

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

Kudos [?]: 395 [0], given: 40

Intern
Joined: 12 Dec 2016
Posts: 1

Kudos [?]: [0], given: 4

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

Kudos [?]: [0], given: 4

Senior Manager
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 439

Kudos [?]: 168 [0], given: 182

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

Kudos [?]: 168 [0], given: 182

Re: On every game, there are only two cases: the player wins or loses   [#permalink] 18 Aug 2017, 07:16
Display posts from previous: Sort by