In a jar there are 3 red balls and 2 blue balls. What is the

Question

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

Bunuel · Answer

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

Answer: A.

Bunuel · Answer

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 in-a-drawer-of-shirts-8-are-blue-6-are-green-and-4-are-24418.html

LucyDang · Answer

Hi Bunuel,

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

LucyDang · Answer

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} .

sshrivats · Answer

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 · Answer

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 · Answer

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 · Answer

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 · Answer

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

CounterSniper · Answer

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

longhaul123 · Answer

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 · Answer

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

TimeTraveller · Answer

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.

pandajee · Answer

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 · Answer

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.

bumpbot · Answer

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