In a jar there are 3 red balls and 2 blue balls. What is the : Problem Solving (PS)

LucyDang

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31 Jul 2014, 07:33

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Bunuel

LucyDang

In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?

A. 9/10
B. 16/20
C. 2/5
D. 3/5
E. ½

P(at least one red) = 1 - P(no red, so 2 blue) = 1- 2/5*1/4 = 9/10.

Answer: A.

Some "at least" probability questions to practice:
leila-is-playing-a-carnival-game-in-which-she-is-given-140018.html
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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
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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
in-a-drawer-of-shirts-8-are-blue-6-are-green-and-4-are-24418.html

Hi Bunuel,

You're super fast!!! Many thanks to your resources! Indeed, I'm practicing probability right now, hihi..

LucyDang

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31 Jul 2014, 20:40

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Another way to solve this problem is:

P (at least 1 red) = 1 - P (no red)

$P(H=2)=\frac{C^2_2}{C^2_5}=\frac{1}{10}$

Hence, $P({at least 1 red})=1-\frac{1}{10}=\frac{9}{10}$.

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31 Jul 2014, 21:15

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P ( at least one red) = 1 - P ( all blue balls )

P( all blue balls ) = 2/5 ( for the first draw) * 1/4 ( for the second draw)
= 2/20 = 1/10

P ( at least one red ) = 1 - 1/10 = 9/10

gooner

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05 Aug 2014, 18:57

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I see that using $P(at least 1 red)$ = $1 - P(no red)$ is the quick way to solve this.

However, should $P(at least 1 red) = P(1 red) + P(2 red)$ work?

where $P(2 red) = \frac{3C2}{5C2}$ and $P(1 red)= \frac{2(3C1*2C1)}{5C2}$

Its not working ofcource because $P(1 red)= \frac{2(3C1*2C1)}{5C2}$ is greater than 1. I multiplied by 2 because I think I have seen Bunuel solve such problems and account for the two ways of picking two balls i.e (B,R) or (R,B)

Where am I wrong in my thinking?

Thanks for the help

Sowmith

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05 Aug 2014, 19:33

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gooner

I see that using $P(at least 1 red)$ = $1 - P(no red)$ is the quick way to solve this.

However, should $P(at least 1 red) = P(1 red) + P(2 red)$ work?

where $P(2 red) = \frac{3C2}{5C2}$ and $P(1 red)= \frac{2(3C1*2C1)}{5C2}$

Its not working ofcource because $P(1 red)= \frac{2(3C1*2C1)}{5C2}$ is greater than 1. I multiplied by 2 because I think I have seen Bunuel solve such problems and account for the two ways of picking two balls i.e (B,R) or (R,B)

Where am I wrong in my thinking?

Thanks for the help

Hi Gooner,

Here order of choosing Blue and Red do not matter. Hence do not multiply with 2.
$P(1 red)= \frac{(3C1*2C1)}{5C2}$
$P(2 red)= \frac{(3C2)}{5C2}$

Finally probability of choosing atleast one red is P(1 red) + p(2 red) = $\frac{(3C1*2C1)}{5C2}$ + $\frac{(3C2)}{5C2}$.

i.e., 3+6/10 = 9/10.

gooner

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05 Aug 2014, 19:55

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Sowmith

gooner

I see that using $P(at least 1 red)$ = $1 - P(no red)$ is the quick way to solve this.

However, should $P(at least 1 red) = P(1 red) + P(2 red)$ work?

where $P(2 red) = \frac{3C2}{5C2}$ and $P(1 red)= \frac{2(3C1*2C1)}{5C2}$

Its not working ofcource because $P(1 red)= \frac{2(3C1*2C1)}{5C2}$ is greater than 1. I multiplied by 2 because I think I have seen Bunuel solve such problems and account for the two ways of picking two balls i.e (B,R) or (R,B)

Where am I wrong in my thinking?

Thanks for the help

Hi Gooner,

Here order of choosing Blue and Red do not matter. Hence do not multiply with 2.
$P(1 red)= \frac{(3C1*2C1)}{5C2}$
$P(2 red)= \frac{(3C2)}{5C2}$

Finally probability of choosing atleast one red is P(1 red) + p(2 red) = $\frac{(3C1*2C1)}{5C2}$ + $\frac{(3C2)}{5C2}$.

i.e., 3+6/10 = 9/10.

Thanks for that. I am trying to think of the scenario where I saw the solution account for the two ways of selecting the desired outcome.

Ashishsteag

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13 Jul 2016, 03:55

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3 Red and 2 Blue balls

Balls taken out can be RR or RB

RR=Number of combinations=2!/2!
RB=Number of combinations=2!(RB or BR)

P(RB)+P(RR)=2*3/5*2/4+3/5*2/4=9/10

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13 Jul 2016, 08:19

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p(no red balls drawn)=2/5*1/4=1/10
p(at least 1 red)=1-1/10=9/10

longhaul123

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09 Jul 2017, 17:47

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In a jar there are 3 red balls and 2 blue balls. What is the probability og drawing at least one red ball when drawing two consecutive balls randomly.
A.9/10
B.16/20
C.2/5
D.3/5
E.1/2

Why are we considering this question to be without replacement ?? Can someone please explain under what conditions do we have to do so? Thank you

rohit8865

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09 Jul 2017, 19:00

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longhaul123

In a jar there are 3 red balls and 2 blue balls. What is the probability og drawing at least one red ball when drawing two consecutive balls randomly.
A.9/10
B.16/20
C.2/5
D.3/5
E.1/2

Why are we considering this question to be without replacement ?? Can someone please explain under what conditions do we have to do so? Thank you

Because he draws two consecutive balls randomly. therefore with replacement could not be considered...

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09 Jul 2017, 19:08

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Usually if the ball is replaced it will either be mentioned specifically, or you will be given that the ball was drawn "one by one" or "one after another", instead of consecutively.

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24 Nov 2019, 00:57

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Bunuel

LucyDang

In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?

A. 9/10
B. 16/20
C. 2/5
D. 3/5
E. ½

P(at least one red) = 1 - P(no red, so 2 blue) = 1- 2/5*1/4 = 9/10.

Answer: A.

Bunuel

I understood that the best way to solve these "Atleast" problem is to find ( 1 - P) way.
But i proceeded by actually finding the Probabilities as given below:
Probability of drawing both red balls = 3/5 x 2/4 =6/20 = 3/10
Probability of drawing one red & one blue ball= 3/5 x 2/4 = 3/10 ( even if we draw blue ball first, the final value remains the same. = 2/5 x 3/4 =6/20 = 3/10)

Now adding both the probabilities, 3/10 + 3/10 = 2 x 3/10 = 6 /10 = 3/5 which is answer D

Can you tell where i went wrong.

Bunuel

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24 Nov 2019, 02:04

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pandajee

Bunuel

LucyDang

In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?

A. 9/10
B. 16/20
C. 2/5
D. 3/5
E. ½

P(at least one red) = 1 - P(no red, so 2 blue) = 1- 2/5*1/4 = 9/10.

Answer: A.

Bunuel

I understood that the best way to solve these "Atleast" problem is to find ( 1 - P) way.
But i proceeded by actually finding the Probabilities as given below:
Probability of drawing both red balls = 3/5 x 2/4 =6/20 = 3/10
Probability of drawing one red & one blue ball= 3/5 x 2/4 = 3/10 ( even if we draw blue ball first, the final value remains the same. = 2/5 x 3/4 =6/20 = 3/10)

Now adding both the probabilities, 3/10 + 3/10 = 2 x 3/10 = 6 /10 = 3/5 which is answer D

Can you tell where i went wrong.

P(at least one red) = P(1 red, 1 blue) + P(2 red) = 3/5*2/4*2 + 3/5*2/4 = 9/10. You should multiply 3/5*2/4 by 2 because 1 red and 1 blue can occur in two ways: RB or BR.

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