Sample space and bridge game

  • MHB
  • Thread starter WMDhamnekar
  • Start date
  • #1
WMDhamnekar
MHB
359
28
Bridge : For k= 1,2,3 ,4 let $N_k$ be the event that North has at least k aces. Let $S_k, E_k, W_k$ be be the analogous events for South, East, West. Discuss the number x of aces in West's possession in the events
a)$W_1', $
b) $N_2S_2,$

c) $N_1'S_1'E_1'$
d) $W_2- W_3$
e)$N_1S_1E_1W_1$

f) $N_3 W_1$
g)$(N_2 \cup S_2)E_2$

How to answer these questions? Truly speaking, I didn't follow these events. Would any member of math help board answer these questions along with explanation of these events?
 
Last edited:

Answers and Replies

  • #2
Kansas Boy
45
0
It helps to know the symbolism. In symbolic logic the prime, ', indicates that the statement is NOT true. Since we are told that \(\displaystyle W_1\) means "West has at least one ace", \(\displaystyle W_1'\) means "West does NOT have at least one ace". I.e. West does not have any aces.

Symilarly, two symbols written together indicate that they are both true. \(\displaystyle N_2S_2\) means that "North has two at least aces and South has at least two aces".

The union sign, \(\displaystyle \cup\), indicates "either or". \(\displaystyle (N_2\cup S_2)E_2\) means "Either North has at least two aces or South has at least two aces, and East has at least two aces.

Finally A- B means that statement A is true but statement B is false. \(\displaystyle W_2- W_3\) means "West has at least two aces but West does not have at least three aces". I.e., West has exactly three aces.

Now you try the others.
 
  • #3
Kansas Boy
45
0
Finally A- B means that statement A is true but statement B is false. \(\displaystyle W_2- W_3\) means "West has at least two aces but West does not have at least three aces". I.e., West has exactly three aces.

The last sentence should have been "West has exactly TWO aces".
 
  • #4
WMDhamnekar
MHB
359
28
The last sentence should have been "West has exactly TWO aces".
The number x of aces in West's possession in the event $(a)W_1', (b)N_2S_2, (g)(N_2 \cup S_2)E_2$ is = 0.

The number of aces in West's possession in the event $(e) N_1S_1E_1W_1, (f)N_3W_1$ is = 1.

The number of aces in West's possession in the event $(d)W_2- W_3$ is = 2.

The number of aces in West's possession in the event $(c) N_1' S_1'E_1'$ is = 4.
 

Suggested for: Sample space and bridge game

  • Last Post
Replies
1
Views
929
  • Last Post
Replies
0
Views
366
Replies
1
Views
421
  • Last Post
Replies
2
Views
458
Replies
2
Views
419
Replies
2
Views
369
Replies
17
Views
635
MHB Math game
  • Last Post
Replies
0
Views
743
  • Last Post
Replies
2
Views
356
  • Last Post
Replies
2
Views
603
Top