Hi Bunuel

According to me dr are two ways to solve the question

First lets get the overview

Event 1(A) = Flipping the quarter( Probability of both winning/loosing = 1/2)

Event 2(B) = Rolling a dice( Probability of winning = 1/6 ;

Probability of loosing = 1 - 1\6 =

5\6)

Event 3(C) = Drawing a card( SPADES) Probability of winning = 13/52=3/4 ;

Probability of loosing = 1 - 3/4 =

1/4)

So now as above we have three events A,B & C.

1st method(This is lengthy method)Possible cases in which it's a win( the highlighted green event is a win and red is loose.

1. ABC = 1/2*5/6*3/4

OR

2.ABC = 1/2*1/6*3/4

OR

3.ABC = 1/2*5/6*1/4

OR

4.ABC = 1/2*1/6*3/4

OR

5.ABC = 1/2*5/6*1/4

OR

6.ABC = 1/2*5/6*1/4

OR

7.ABC = 1/2*1/6*1/4

As we now OR means add

Adding up all of them we'll get the probability of winning ie 11/16

2nd Method (For those whose who like short methods) dr's a concept that Probability of not happening = 1 - Probability of happening

This is what we gonna use here

No let's calculate Probability of loosing in all the three events

P of loosing in A = 1 - 1/2 = 1/2

P of loosing in B = 1 - 1/6 = 5/6

P of loosing in C = 1 - 1/4 = 3/4

Therefore P of loosing the game = (P of loosing in A) * (P of loosing in B) * (P of loosing in C) = 1/2*5/6*3/4 = 5/16

Now P of winning = 1 - 5/16 = 11/16

_________________

Thank you

+KUDOS

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