How Do Ace Distributions Affect Probabilities in a Bridge Game?

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WMDhamnekar
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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?
 
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It helps to know the symbolism. In symbolic logic the prime, ', indicates that the statement is NOT true. Since we are told that [math]W_1[/math] means "West has at least one ace", [math]W_1'[/math] 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. [math]N_2S_2[/math] means that "North has two at least aces and South has at least two aces".

The union sign, [math]\cup[/math], indicates "either or". [math](N_2\cup S_2)E_2[/math] 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. [math]W_2- W_3[/math] 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.
 
County Boy said:
Finally A- B means that statement A is true but statement B is false. [math]W_2- W_3[/math] 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".
 
County Boy said:
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.