Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If three balls are randomly selected from a collection of 20 red, 12 blue, and 9 white balls, assuming the balls are identical except for the difference in colors, what is the probability of select

If three balls are randomly selected from a collection of 20 red, 12 blue, and 9 white balls, assuming the balls are identical except for the difference in colors, what is the probability of selecting a white ball and a blue ball on the first two draws, and a red ball on the third draw?

 

Option: 1

0.02


Option: 2

0.07


Option: 3

0.11


Option: 4

0.3


Answers (1)

best_answer

To find the probability of selecting a white ball and a blue ball on the first two draws, and a red ball on the third draw, we need to calculate the probability of each possible outcome.

The probability of selecting a white ball on the first draw is given by the ratio of the number of white balls to the total number of balls:

P(White on first draw) = (number of white balls) / (total number of balls)

After the first draw, there will be 8 white balls remaining, out of which 12 are blue. The probability of selecting a blue ball on the second draw is given by the ratio of the number of blue balls to the total number of balls remaining:

P(Blue on second draw) = (number of blue balls) / (total number of balls remaining)

After the second draw, there will be 11 balls remaining, out of which 20 are red. The probability of selecting a red ball on the third draw is given by the ratio of the number of red balls to the total number of balls remaining:

P(Red on third draw) = (number of red balls) / (total number of balls remaining)

To find the overall probability of selecting a white ball and a blue ball on the first two draws, and a red ball on the third draw, we multiply the probabilities of each individual draw:

Overall probability = P(White on first draw) \times P(Blue on second draw) \times P(Red on third draw)

Let's calculate the values:
\mathrm{P(\text { White on first draw })=9 /(20+12+9)=9 / 41}
\mathrm{ P(\text { Blue on second draw })=12 /(19+11+8)=12 / 38 }
\mathrm{ P(\text { Red on third draw })=20 /(18+10+7)=20 / 35=4 / 7}
\mathrm{ \text { Overall probability }=(9 / 41) \times(12 / 38) \times(4 / 7) \approx 0.0217}

Therefore, the probability of selecting a white ball and a blue ball on the first two draws, and a red ball on the third draw is approximately 0.0217 , or 2.17 \%.

 

Posted by

Suraj Bhandari

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 registration last day, city slip soon; here's how to prepare
anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow

Latest Question