Some "at least" probability questions to practice:
leila-is-playing-a-carnival-game-in-which-she-is-given-140018.html
a-fair-coin-is-tossed-4-times-what-is-the-probability-of-131592.html
for-each-player-s-turn-in-a-certain-board-game-a-card-is-132074.html
a-string-of-10-light-bulbs-is-wired-in-such-a-way-that-if-131205.html
a-shipment-of-8-tv-sets-contains-2-black-and-white-sets-and-53338.html
on-a-shelf-there-are-6-hardback-books-and-2-paperback-book-135122.html
in-a-group-with-800-people-136839.html
the-probability-of-a-man-hitting-a-bulls-eye-in-one-fire-is-136935.html
for-each-player-s-turn-in-a-certain-board-game-a-card-is-141790.html
the-probability-that-a-convenience-store-has-cans-of-iced-128689.html
triplets-adam-bruce-and-charlie-enter-a-triathlon-if-132688.html
a-manufacturer-is-using-glass-as-the-surface-144642.html
the-probability-is-1-2-that-a-certain-coin-will-turn-up-head-144730.html (OG13)
a-fair-coin-is-to-be-tossed-twice-and-an-integer-is-to-be-148779.html
in-a-game-one-player-throws-two-fair-six-sided-die-at-the-151956.html
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1) 4/10 * 3/9 * 2/8 = 2/5 * 1/3 * 1/4 = 2/60 = 1/30

2) 6/10 * 5/9 * 4/8 = 3/5 * 5/9 * 1/2 = 3/18 = 1/6

3) 1 - P(all boys selected) = 1- (1/6) = 5/6

1) (4/10)(3/9)(2/8) = 4/105
2) (6/10)(5/9)(4/8) = 4/21
3) (4/10)(6/9)(5/8) = (1/6)*3 = 1/2

I must be stupid. I was wondering how I got it wrong so I went back to my numbers and somehow I multiplied by 7 when I had 8:
4/10)(3/9)(2/8 but mutliplied by 7) = 4/105 ... WTF?

Is third right, though? 1-all boys possibility?

very nice. all are correct

for #3, we can also do (10C3 - 6C3)/10C3 = 100/120 = 5/6

4/10*3/9*2/8 24/720 --> 12/360 --> 6/180 --> 3/90 --> 1/30

6/10*5/9*4/8 --> 120/720--> 12/72 --> 1/6

1-Probability that none are girls: --> just use the above 1-6/10*5/9*4/8 = 5/6.

For 3Q. What is the probability that at least one girl will be elected?

Isn't it.... 1-p(no girls selected) = 1-(6/10) = 4/10 = 2/5


Can someone give, y above is wrong..pls

1 - p(no girls selected) is the correct strategy, p(no girls selected) isn't as simple as 6/10

6/10 is the probability of one boy to be "pulled out of the hat", the next boy is 5/9 and so forth

p(no girls selected) is therefore p(all boys selected) = 6C3/10C3 = 1/6
as shown above

I believe this is a permuation problem as the order does matter...

What is the probability that only girls will be elected? would be 4p3/10p3 = 1/30

Its a coincidence that the answer is 1/30 when combinations is used i.e., 4c3/10c3.

1.4C3/10C3
2. 6C3/10C3
3. 1-6C3/10C3

I have another opinion on the third question. Wonder, if it is right?

1) I got 1/30 too
2) got 1/6 as well

But, in my opinion, the third question should be solved this way:
the question asks what is the probability of AT LEAST one girl will be selected.

we have three mutually exclusive possibilities to select (they are all satisfy the asked condition):
GBB (girl,boy,boy)
GGB (girl,girl,boy)
GGG (girl,girl,girl)

the probability of GGG is already known - 1/30.
the probability of GGB is: 4/10*3/9*6/8 - 1/10.
the probability of GBB is: 4/10*6/9*5/8 - 1/6.

As these are mutually exclusive, we shall add them together: 1/30+1/10+1/6 = 3/10.

What do you think of this?

ANS.......
u r right but only mistake u r doing is that u have to multiply two of possiblities by 3....reason:-
possiblities of taking two girls is GGB,GBG,BGG ie 1/10*3=3/10.....
possiblities of taking two girls is GBB,BGB,BBG ie 1/6*3=1/2....
As these are mutually exclusive, we shall add them together: 1/30+3/10+1/2 = 5/6.
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Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

What is the probability that at least one girl will be elected?

Soln:
All are permutations since even though we select 3 people , they are pushed into different posts. i.e. We could have selected the same 3 people, and arrange them in 6 different ways (3!) coz the posts that they take are different. president ,vice president and secretary.

What is the probability that only girls will be elected?
Ans = 4P3/10P3

What is the probability that only boys will be elected?
Ans = 6P3/10P3

What is the probability that at least one girl will be elected?
Ans = (4P1 * 6P2 + 4P2 * 6P1 + 4P3)/10P3

Yup, this is a permutation problem since three people will be selected for three different posts.

A) 4P3/10P3

B) 6P3/10P3

C) 1- (6P3/10P3)

I got the answers right but have a doubt.

If the teacher selects for president, vice pres & secretary, there are 3 distinct positions unlike say getting selected for a generic team where it doesn't matter what kind of position.

So girl 1 can be selected for pres, girl 2 for vp & girl 3 for secretary is a different case than girl 3 for pres, girl 2 for vp & girl 1 for secretary.

Can someone explain why we dont consider them distinct possibilities in this case?
thanks.

IMO.. it shouldn't matter whether it is calculated with combinations or permutations. Reason being, if you consider the order in choosing the girls (numerator), you do the same in the denominator too. Hence it balances it out...

the results are coming out to be the same....

Cheers!
JT

Total comb = 10c3

1. Only girls = 4c3/10c3 = 1/30
2. Only boys = 6c3/10c3 = 1/6
3. Atleast one gurl = 1 - no gurl = 1 - only boys = 1 - 1/6 = 5/6

Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

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(1) What is the probability that only girls will be elected?

Just for practice, since the most efficient way to solve this problem is the Probability approach.

(1) probability approach:

\(\frac{4}{10} * \frac{3}{9} * \frac{2}{8} = \frac{1}{30}\)

But I need to practice so I solved this problem with the other 3 approaches:

(2) Reversal Probability approach:
P = 1 - q (q=probability of not picking only girls, this means, picking 1, 2 or 3 boys):

--> 1 boy + 2 girls = BGG (and combinations):
\(1 - \frac{6}{10} * \frac{4}{9} * \frac{3}{8} * \frac{3!}{2!} = \frac{3}{10}\)

-->2 boys + 1 Girl = BBG (And combinations):
\(\frac{6}{10} *\frac{5}{9} * \frac{4}{8} * \frac{3!}{2!} = \frac{1}{2}\)

--> 3 boys:
\(\frac{6}{10} * \frac{5}{9} * \frac{4}{8} * \frac{3!}{3!} = \frac{1}{6}\)

Then, we add all the probabilities:

\(\frac{3}{10} + \frac{1}{2}+ \frac{1}{6} = \frac{29}{30} = q\)

\(1-q = 1- \frac{29}{30} = \frac{1}{30}\) = P

(3) Combinatorial approach:

\(\frac{C^4_3}{C^10_3} = \frac{4}{120} = \frac{1}{30}\)

(4) Reversal combinatorial approach:

\(1 - \frac{{C^6_1 * C^4_2 + C^6_2 * C^4_1 + C^6_3}}{C^10_3} = 1- \frac{{36 + 60 + 20}}{120} = 120/120 - \frac{116}{120} = \frac{1}{30}\)

(2) What is the probability that only boys will be elected?

(1) Probability approach:

\(\frac{6}{10} * \frac{5}{9} * \frac{4}{8} = \frac{4}{24} = \frac{1}{6}\)

(2) Reversal probability approach --> P = 1 - q

1girl + 2 boys: \(\frac{4}{10} *\frac{6}{9} *\frac{5}{8} * \frac{3!}{2!} = \frac{1}{6} * 3 = \frac{1}{2}\)

2 girls + 1 boy: \(\frac{4}{10} *\frac{3}{9} * \frac{6}{8} *\frac{3!}{2!} = \frac{1}{10} * 3 = \frac{3}{10}\)

3 girls: \(\frac{4}{10} *\frac{3}{9} * \frac{2}{8} = \frac{1}{30}\)

(3) Combinatorial approach:

\(\frac{C^6_3}{C^10_3} = \frac{20}{120} = \frac{1}{6}\)

(4) Reversal Comb approach:

\(\frac{C^4_1 * C^6_2 + C^4_2 * C^6_1 + C^4_3}{C^10_3}\)

1 girl + 2 boys: \(C^4_1 * C^6_2 = 60\)
2 girls + 1 boy: \(C^4_2 * C^6_1 = 36\)
3 girls:\(C^4_3 = 4\)

60 + 36 + 4 = 100

\(\frac{120}{120} - \frac{100}{120} = \frac{20}{120} = \frac{1}{6}\)

(1). All girls are selected.

Just for argument sake, it has not been mentioned that One GIRL/BOY cannot be absorbed in multiple roles.

So why the probability P(All Girls) is not calculated as:

(4*4*4)/(10*10*10) = (8/125)

Desired outcome= You can fill each position in 4 ways or select each position in 4 ways
Total outcome = You can make 10 selections each time

Please advise !!

Rgds,
TGC !