Joshua and Jose work at an auto repair center with 4 other

P(Joshua and Jose both chosen) = P(one of the two men is chosen first AND the other man is chosen second)
= P(one of the two men is chosen first) x P(the other man is chosen second)
= 2/6 x 1/5
= 1/15

Answer: A

Cheers,
Brent
_________________

Two Methods

1) Probability of chosing Josh first = 1/6
Probability of chosing Jose second = 1/5
total = 1/30
Probability of chosing Jose first = 1/6
Probability of chosing Josh second = 1/5
Total = 1/30
Final = 1/30 + 1/30 = 1/15

2) Number of ways two persons can be chosen 6C2 = 15
Number of ways Josh and Jose are the two persons = 1
Total = 1/15

We should consider Joshua and Jose as one entity and then proceed for the probability.

Total ways of choosing 2 people (Joshua and Jose) out of 6 people --
\((6!)/(4! * 2!)\)
\(=15\)

Favorable outcomes - 1 (Since Joshua and Josh should be picked up)

Probability - \(1/15\)

Any other thoughts.....

Interestingly in such questions if one notices the denominator, we know it has to be 15, as 6C2=15, so that rules out B C D. Leaves us with A and B. A says 1/15, B says 5/15. of course there is one choice and that is 1/15, but even if you did not get it the right way, this elimination should help.

I reworded this problem to have Josh and Jose be two guys, and the other 4 are women. What are the chances of picking only guys?

There are two slots, first slot has (2/6) chance of picking a guy, the second slot has (1/5) chance of picking the other guy.

Picture them in a group. The first time you randomly pick someone, there are six people, and two guys. Now, assuming you have picked a guy, you are left with a group of 5 people standing there (1 guy, 4 girls). The chance of picking a guy then is 1/5. Multiply the two probabilities together, and you have 2/30, or 1/15

Probability of an event = (number of positive outcomes)/(total number of outcomes)

Here number of positive outcomes = choose Josh and Jose when two people are picked at random = 2C2

Total Outcomes = choose two people from a group of 6 people = 6C2

Probability = 2C2 / 6C2 = 1/15

total ways to select 2 people out of 6 = 6c2 = 15
the number of ways to select J & J = 2c2 = 1

= 1/ 15 (answer)

P (picking Joshua) first is 1/6 and picking Ann second 1/5.

1 st case So, P(J & A)=1/5 x 1/6 = 1/30 - J. and Ann second.

and in the reverse order,

2nd case P(A & J) = 1/30- Ann first, Joshua second

Add them = 1/15.

Can someone spot me the error in this alternate method,

total outcome: 6c2
non favourable outcome:4c2 ( leaving out josha and jose)

req prob -> 1- (4c2/6c2)
reduces to 9/15.!!

Hi,

Non Fav o/comes =
Leaving out Joshua and Jose +
Leaving out Joshua and including Jose +
Leaving out Jose and including Joshua

Leaving out Joshua and Jose = 4C2
Leaving out Joshua and including Jose = 4C1 (as Jose has already been selected)
Leaving out Jose and including Joshua = 4C1 (as Joshua has already been selected)

When these cases are eliminated, one does get the answer of 1/15.

Hope this helps. Thanks.

Total ways of choosing two people out of 6 = 6c2=15.
And out of these 15 cases there is only one case in which Joshua and Jose both are selected. hence 1/15 is the probability.

The other 14 cases would be : When Joshua is selected and one person out of the other four (excluding Jose) is selected. 1x4c1=4 cases. When Jose is selected and one person out of the other four (excluding Joshua) is selected. 1x4c1=4 cases. When none of Joshua and Jose is selected. 4c2=6 cases. This adds up to 4+4+6=14 cases.

I am not able to solve this questions with below method:

probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5

probability required = 1-2/5 = 3/5

Where am I doing wrong?

Regards,
Ravi

You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.
_________________

Hmmm...silly again.

Only Joshua but not jose probability = 4/15
only jose but not joshua = 4/15

2/5 +4/15+4/15= 14/15

so actual probabilty required is 1-14/15 = 1/15

Cheers Bunuel!!

Bunuel

I went for a simpler approach and although I get the right answer, I'm not sure if it is correct. Here is how I went about it:

Since the question is only concerned about Josh and Jose, the probability that either one of them is chosen first is: 2/6
After the first is chosen, there are 5 people left, out if which one is either Josh or Jose. Probability of getting them is then: 1/5

So, (2/6)*(1/5)= 1/15

Is this a right approach for this question? To me this seems so much more straightforward...Thoughts? Suggestions?

Yes, that's correct.
_________________

I am not able to solve this questions with below method:

probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5

probability required = 1-2/5 = 3/5

Where am I doing wrong?

Regards,
Ravi[/quote]

You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.[/quote]


Hmmm...silly again.

Only Joshua but not jose probability = 4/15
only jose but not joshua = 4/15

2/5 +4/15+4/15= 14/15

so actual probabilty required is 1-14/15 = 1/15

Cheers Bunuel!![/quote]
I'm afraid I don't get the red part: only joshua and not jose: 1/6 * 4/5 = 4/30 =2/15. Do we now multiply by 2 because you can pick them in two ways (pick anyone but jose first and pick joshua second) ?

You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.[/quote]


Hmmm...silly again.

Only Joshua but not jose probability = 4/15
only jose but not joshua = 4/15

2/5 +4/15+4/15= 14/15

so actual probabilty required is 1-14/15 = 1/15

Cheers Bunuel!![/quote]
I'm afraid I don't get the red part: only joshua and not jose: 1/6 * 4/5 = 4/30 =2/15. Do we now multiply by 2 because you can pick them in two ways (pick anyone but jose first and pick joshua second) ?[/quote]

=====================
A> Only Joshua(Jh) and !Jose(Jo)

so we have Joshua + Any of the (A,B,C,D)= (Jh,A) or (Jh,B) or (Jh,C) or (Jh,D) ------>[or simply we can write it as 4c1 since we have Jh already and we have to choose one from other 4]

4c1/6c2

B> Only Jo and !Jh

same 4c1/6c2

final answer =

1- (probabilty of not choosing both of them)- probability of choosing Jh but !Jo - Probability of choosing Jo but !Jh.


This is bulky and I was just checking if things would work this way...simple way is

there can only be one set with both Jo and Jh and total number of ways choosing 2 from 6 is 6c2 = 15

therefore probability is 1/15

Hmmm...silly again.

Only Joshua but not jose probability = 4/15
only jose but not joshua = 4/15

2/5 +4/15+4/15= 14/15

so actual probabilty required is 1-14/15 = 1/15

Cheers Bunuel!![/quote]
I'm afraid I don't get the red part: only joshua and not jose: 1/6 * 4/5 = 4/30 =2/15. Do we now multiply by 2 because you can pick them in two ways (pick anyone but jose first and pick joshua second) ?[/quote]

=====================
A> Only Joshua(Jh) and !Jose(Jo)

so we have Joshua + Any of the (A,B,C,D)= (Jh,A) or (Jh,B) or (Jh,C) or (Jh,D) ------>[or simply we can write it as 4c1 since we have Jh already and we have to choose one from other 4]

4c1/6c2

B> Only Jo and !Jh

same 4c1/6c2

final answer =

1- (probabilty of not choosing both of them)- probability of choosing Jh but !Jo - Probability of choosing Jo but !Jh.


This is bulky and I was just checking if things would work this way...simple way is

there can only be one set with both Jo and Jh and total number of ways choosing 2 from 6 is 6c2 = 15

therefore probability is 1/15 [/quote]

Got it thanks. but what is wrong with my logic? if you could help, that'll be great. It'll help me conceptually.