Finding a Mistake: Probability of Y with X and Spins

[songoku]

Homework Statement

A fair five-sided spinner is numbered 3, 4, 5,6, and 7. The spinner is spun and the number showing, X, is recorded. The spinner is spun until it gets an odd number or until it has been spun 3 times. The number of spins taken to get an odd number, Y, is recorded.

Find P(X > Y + 3)

Relevant Equations

Probability

I divide the question into three cases:

1) P (Y = 1 and X = 5 or 7) = 1/5 + 1/5 = 2/5

2) P(Y = 2 and 1st spin = even and 2nd spin = 7) = 2/5 x 1/5 = 2/25

3) P(Y = 3 and 1st and 2nd spin = even and 3rd spin = 7) = 2/5 x 2/5 x 1/5 = 2/125

Total probability = 2/5 + 2/25 + 2/125 = 62/125

But the answer key is 61/125

Where is my mistake? Thanks

 

[andrewkirk]

Your main mistake is to assume you have made a mistake :-D
Remember that if your answer doesn't match the one in the book, that means at least one of the two must be wrong, but doesn't tell you which one. People that write books make mistakes too.
In fact, I think both your answer and the book's answer contain mistakes, and the correct answer is 64/125, because item (3) above should be 4/125.

 

[songoku]

Thank you very much andrewkirk

 

[WWGD]

Thank you very much andrewkirk

Now find the probability that either your or the book's answer is wrong ;).

 

[WWGD]

These types of problems often have a nice geometrical interpretation, e.g., $P( X >Y +3)$
is the area of $X$ above the line $Y+3$ . This is useful often in, e.g., waiting time problems. Say $X,Y$ are supposed to meet in $[0,1]$ randomly , or according to different conditions. Then we integrate over corresponding region in $[0,1] \times [0,1]$

 

[Gavran]

$\text { There are three answers: } \frac { 61 } { 125 } \text { , } \frac { 62 } { 125 }$

$\text { and } \frac { 64 } { 125 } \text { . It is already said why } \frac { 62 } { 125 }$

$\text { is not a correct answer, but } \frac { 64 } { 125 } \text { is not a correct answer }$

$\text { too. }$$\text { Observing the event } X \gt Y + 3 \text { as a union of the one spin event, the }$

$\text { two spins event and the three spins event is a good approach. Observing two }$

$\text { spins as two independent events is a good approach too. The choice that each }$

$\text { outcome of every single spin occurs with equal probability, which is }$

$\frac { 1 } { 5 } \text { , is a good choice. }$$\text { The number of outcomes, when } X \gt Y + 3 \text { , must be determined }$

$\text { properly and here it is done only for the one spin event when } Y = 1$

$\text { and when the number of outcomes is } 2 \text { . }$

$\text { For the two spins event } X \text { got in the first spin must be greater than }$

$Y + 3 = 5 \text { and } X \text { got in the second spin must be greater than } 5$

$\text { too. This is why outcome } ( 4, 7 ) \text { can not be included into the two }$

$\text { spins event. }$

$\text { Now it is easy to make judgement if the three spins event when } Y = 3$

$\text { and when } X \gt Y + 3 \text { is impossible event or not. }$$\text { The answer } \frac { 61 } { 125 } \text { can only be correct if } X \geq Y + 3$

$\text { , what can be checked. }$

 

[haruspex]

The question as quoted is ambiguous. Is the first spin, producing X, also the first of the three? I assume not.
Since Y is at least 1, we only have to consider X=5, 6 or 7. The probabilities for the number of spins are
Y
1: 15/25
2: 6/25
3: 4/25
So we have to add:
Y
1: 1x1/5x15/25 (X=5)
2: 2x1/5x6/25 (X=5,6)
3: 3x1/5x4/25 (X=5,6,7)
to get 61/125.

 

[Gavran]

$\text { Is the question really ambiguous or being interpreted wrongly, causing }$

$\text { a lot of errors? }$

$\text { There are two events, the first one, which consists of only one spin, gives } X$

$\text { and the second one, which consists of one spin, two spins or three spins, gives }$

$Y \text { . This interpretation is only acceptable. }$

$\text { Probability values } P ( Y = 1 ) \text { , } P ( Y = 2 ) \text { and } P ( Y = 3 )$

$\text { are determined well. }$

$\text { The expression }$

$1 \cdot \frac 1 5 \cdot \frac { 15 } { 25 } + 2 \cdot \frac 1 5 \cdot \frac { 6 } { 25 } + 3 \cdot \frac 1 5 \cdot \frac { 4 } { 25 }$

$\text { gives } \frac { 39 } { 125 } \text { , which is not a correct probability value of }$

$\text { the event } X \gt Y + 3 \text { . Associated values of } X \text { with particular }$

$\text { values of } Y \text { are determined well only for } Y = 2 \text { where }$

$X \in \left \{ 6 , 7 \right \} \text { . } X = 5 \text { or } X = 6 \text { can not be }$

$\text { associated with } Y = 3 \text { because } X \text { is greater than } Y + 3$

$\text { . Now values of } X \text { associated with } Y=1 \text { can be determined }$

$\text { easily. }$

$\text { After correcting mistakes, the calculated value of } P ( X \gt Y + 3 ) \text { will }$

$\text { be the correct probability value of the event } X \gt Y + 3 \text { . }$