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If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive?

Guest May 21, 2014

Best Answer 

 #16
avatar+889 
+8

There are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.)

The counting method has been used for both of the correct answers so far,  so here's the probability method.

Suppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340.

Bertie  May 21, 2014
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21+0 Answers

 #1
avatar+996 
0

Well, three skiiers on a plane of 17. Considering that one person dies each time, the number decreases by 1 in both the numerator and the denominator.

 

$$\left({\frac{{\mathtt{3}}}{{\mathtt{17}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{16}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{15}}}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{680}}}} = {\mathtt{0.001\: \!470\: \!588\: \!235\: \!294\: \!1}}$$

 

The odds are 1/680

GoldenLeaf  May 21, 2014
 #2
avatar+92194 
+8

Probability is not my strong suit but this is what i think.

P= number of ways 4 can be chosen from 14 / number of ways 4 can be chosen from 17

P= 14C4 / 17C4

P = 1001 / 2380

$$\mbox{P(all 3 survive) }=\frac{1001}{2380}=\frac{143}{340}$$

$${\frac{{\mathtt{1\,001}}}{{\mathtt{2\,380}}}} = {\frac{{\mathtt{143}}}{{\mathtt{340}}}} = {\mathtt{0.420\: \!588\: \!235\: \!294\: \!117\: \!6}}$$

 

Can another mathematician please check this - I'm fairly confident that it is correct.

Melody  May 21, 2014
 #3
avatar+85644 
+8

Probabilty isn't my strong point, but I'll take a run at this one.

So, basically, we first want to count the sets of people who might die - what a morbid problem!!

The total number of people who could die is given by choosing some group of 4 from the 17 =  C(17,4) = 2380

Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(13,3) = 858 sets

Now, let's look at the possible sets where 2 skiers die - C(3,2) *C(13,2) = 234 sets

Now let's consider the sets where all three skiers die = C(3,3) * C (13,1) = 13

So, the total number of sets containing any of the skiers = 1105

Thus, the chances that all three survive are given by:

1 - (the number of sets containing any of the skiers)/(the total number of possible sets)

=  1 - (858 + 234 + 13)/2380 ≈ 53.6 %

OOPS !!    Melody pointed out a math error  I made....let me correct this...and as she indicated recently, she DOES believe in "do-overs"....I still like my logic, though!

 

Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(14,3) = 1092 sets

Now, let's look at the possible sets where 2 skiers die - C(3,2) *C(14,2) = 273 sets

Now let's consider the sets where all three skiers die = C(3,3) * C (14,1) = 14

So, the total number of sets containing any of the skiers = 1379

Thus, the chances that all three survive are given by:

1 - (the number of sets containing any of the skiers)/(the total number of possible sets)

==  1 - (1379)/2380 ≈ 42.1 %

Thanx, Melody!!

CPhill  May 21, 2014
 #4
avatar+92194 
0

Okay, we have 3 answers - all different 

We need an arbitrator - preferably one who knows what he/she is talking about.

also if possible please explain what is wrong with our logic.

Thank you.

Melody  May 21, 2014
 #5
avatar+92194 
+8

Okay Chris, 

I am taking a look at yours.

Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(13,3) = 858 sets

shouldn't this be 3C1*14C3 = 3*364=1092 ?

Melody  May 21, 2014
 #6
avatar+85644 
+8

We both got the same result...now, let's figure the probability that 2 board members out of (n) board members could arrive at the same (possibly) correct answer!!

No snide remarks here, please.....

CPhill  May 21, 2014
 #7
avatar+92194 
0

EXCELLENT!!!!!

Melody  May 21, 2014
 #8
avatar+85644 
0

HAHA!!!...and 57 minutes ago...you were looking for someone who knew what they were doing!!! (an arbitrator, I believe??)

Are we "sure" our answers are correct??

(Actually....I think they might be)

CPhill  May 21, 2014
 #9
avatar+92194 
0

Yes I am sure.

Because 

Two out of three ain't bad!

Melody  May 21, 2014
 #10
avatar+85644 
0

But 2/3 = 66%

That leaves 1/3 chance = 33% = that we might not be!!

I say...let's call in the "troll" as a referee.....HE KNOWS ALL !!!

LOL!!

CPhill  May 21, 2014
 #11
avatar+92194 
0

Okay

Where's 'our' KNOW-IT-ALL TROLL when we need him?

He's probably AWOL with Sir Cumference!

Melody  May 21, 2014
 #12
avatar+85644 
0

I'm thinking he was on that plane...and maybe he doesn't ski, either......

Mmmmmmm......maybe we've solved a forum "problem"

(At least there's about a 58% chance of it....)

CPhill  May 21, 2014
 #13
avatar+92194 
0

NOW that is a percentage that I would very seriously challenge!

P(troll gone)=P(troll was on plane)*P(troll died) $$\rightarrow 0$$

Melody  May 21, 2014
 #14
avatar+85644 
0

Yeah...we couldn't be that fortunate, huh??

CPhill  May 21, 2014
 #15
avatar+92194 
0

That's not nice Chris.  So long as we keep him on a tight leash he is fun to have around.

You know that just as well as I do!

Melody  May 21, 2014
 #16
avatar+889 
+8
Best Answer

There are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.)

The counting method has been used for both of the correct answers so far,  so here's the probability method.

Suppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340.

Bertie  May 21, 2014
 #17
avatar+2353 
0

But what if the plane was carrying 4 people that died in a crash in coffins in the cargo room. Then the plane might not have crashed at all and the skiers survive 

Unless the skiers died in a skiing accident off course 

reinout-g  May 21, 2014
 #18
avatar
0

If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive?

----

This probability is trickier than it looks.

Four (4) will die and at most only three (3) can be skiers.

By deduction one (1) will die and not be a skier.

Now 16 remain, three (3) of whom are skiers.

From this calculate the probability of selecting (Z) correct out of (R) draws from (N) numbers. (Z) (in this case) defines the probability of a skier dying.

Probability= (R!/(Z!*(R-Z)!) * (N-R)!/(((N-R)-(R-Z))!*(R-Z)!)/(N!/((R!)*((N-R)!)))

 $${Probability =}\frac{\frac{R!}{Z!*(R-Z)!} * \frac{(N-R)!}{((N-R)-(R-Z))!*(R-Z)! }}{\frac{N!}{R!*(N-R)!}}$$

 Column ID’s: A= (R!)/(Z!(R-Z)!)

                    B= (N-R)!/(((N-R)-(R-Z))!*(R-Z)!)

                    C= N!/((R!)*((N-R)!))

                    D= Probability of (Z) skiers dying.

                    E= 1/Probability

N     R     Z     A     B      C                   D                            E

16   3      3     1     1      560     0.001785714285714     560.0000000000

16   3      2     3    13     560     0.069642857142857       14.3589743590

16   3      1     3    78     560     0.417857142857143        2.3931623932

16   3      0     1  286     560     0.510714285714286        1.9580419580

-----------

Probability of zero skiers dying ~ 0.511 (51.1%)

(Professional help provided by Francis and Frances)

by: Someone Who Knows Everything

Guest May 21, 2014
 #19
avatar+92194 
0

deleted deleted

 

Melody  May 21, 2014
 #20
avatar+26625 
0

My view is:

Probability that 4 non-skiers died (hence 3 skiers survived) is:

$${\frac{{\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{13}}{\mathtt{\,\times\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{11}}}{\left({\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{14}}\right)}} = {\frac{{\mathtt{143}}}{{\mathtt{340}}}} = {\mathtt{0.420\: \!588\: \!235\: \!294\: \!117\: \!6}}$$

 

 Oops! Just noticed that this is the same as Bertie's reply (though Bertie did a more complete job by including an explanation).

Alan  May 22, 2014
 #21
avatar+92194 
0

It is the same as mine and Chris's as well!!

Melody  May 22, 2014

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