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AFCAT Numerical Ability · Probability

Drawing from a Bag: One Ball, One After Another, With Replacement

One ball is favourable balls over all balls; draws one after another multiply, with the bag shrinking at each step unless the ball is put back.

All Probability notes

Why this matters

This page is EASY and MODERATE questions only, but each turns on one word: not, or, neither, with replacement or without. Read the bag twice, once for the colours and once for whether a ball goes back, and most of the marks follow.

Concept 1 of 3: One ball: favourable over total

A bag with 3 red, 5 blue and 2 white balls holds 10 balls. One ball drawn at random is blue with probability 510=12\tfrac{5}{10} = \tfrac12.

Definition

  • The total is every ball of every colour. The favourable balls are those of the colours asked for.
  • 'Not red' is all the other colours, or 1−P(red)1 - P(\text{red}).
  • 'Red or white' adds, because one ball cannot be both colours.
  • 'Neither red nor white' is the colours left over.

One ball

P(colour)=balls of that colourall ballsP(\text{colour}) = \dfrac{\text{balls of that colour}}{\text{all balls}}

Worked example

A bag has 7 red, 2 green and 6 white balls. Find the probability that a ball drawn at random is red or green.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Probability · Drawing from a Bag

A bag has 4 green, 5 yellow, and 6 red marbles. If a marble is drawn at random, what is the probability that it is neither green nor red?

The total is every ball

In a bag of 3 red, 5 blue and 2 white, white is 210\tfrac{2}{10}, not 28\tfrac{2}{8}: the white balls are part of the total.

Neither is not either

'Neither red nor blue' is the colours left over. 'Either red or blue' is those two colours. Reading one as the other gives the complement.
Drill 7 more on one ball: favourable over total

Concept 2 of 3: One after another, without replacement

Draw a red ball and keep it out: the bag now has one ball fewer and one red fewer. Multiply the step probabilities, each worked on the bag as it is at that step.

Definition

  • P(first red, then blue) =rn×bn−1= \tfrac{r}{n} \times \tfrac{b}{n - 1}, for r red and b blue among n balls.
  • Both red: rn×r−1n−1\tfrac{r}{n} \times \tfrac{r - 1}{n - 1}.
  • 'The second is red, given the first was blue' is only the second step: rn−1\tfrac{r}{n - 1}.
  • At least one red in two draws: 1−P(no red)1 - P(\text{no red}), where no red is two balls of the other colours, step by step.

Without replacement

P(A then B)=an×bn−1P(A \text{ then } B) = \dfrac{a}{n} \times \dfrac{b}{n - 1}

Worked example

A bag has 5 red and 3 blue balls. Two are drawn one after the other without replacement. Find the probability that the first is blue and the second red.
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Probability · Drawing from a Bag

In a box with 66 red and 44 green balls, what is the probability that in 33 draws without replacement, the first two are red and the third is green?

Change both numbers

After a red ball is kept out of 5 red and 3 blue, the bag has 4 red among 7. Writing 57\tfrac57 or 48\tfrac48 for the next red changes only one of the two.

Order is part of the event

'First red, then blue' is one order. 'One red and one blue' in two draws adds the other order too: 2×58×37=15282 \times \tfrac58 \times \tfrac37 = \tfrac{15}{28}.
Drill 7 more on one after another, without replacement

Concept 3 of 3: With replacement: the bag resets

When each ball goes back before the next draw, every draw is from the same bag. The draws are independent, like tosses of a biased coin.

Definition

  • All k draws red: (rn)k\left(\tfrac{r}{n}\right)^k.
  • Exactly j red in k draws: kCj pj(1−p)k−j^kC_j \, p^j (1 - p)^{k - j}, with p=rnp = \tfrac{r}{n}. It is the coins rule with p in place of a half.
  • One of each of two colours in two draws: add both orders, 2pq2pq, where p and q are the two colours' probabilities.

j of k draws, with replacement

P(j of k)=kCj pj(1−p)k−jP(j \text{ of } k) = {}^kC_j\, p^j (1 - p)^{k - j}

Worked example

A bag has 3 red and 7 white balls. Three are drawn with replacement. Find the probability that exactly two are red.
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Probability · Drawing from a Bag

From a bag containing 99 red and 66 blue balls, what is the probability of drawing 22 balls with replacement such that one is red and one is blue?

With replacement nothing shrinks

Three draws with replacement from 2 red and 8 white give all red as (210)3=1125\left(\tfrac{2}{10}\right)^3 = \tfrac{1}{125}. Writing 210×19×08\tfrac{2}{10} \times \tfrac19 \times \tfrac08 treats the balls as kept out.
Drill 5 more on with replacement: the bag resets

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • One ball: favourable over total

    One ball

    P(colour)=balls of that colourall ballsP(\text{colour}) = \dfrac{\text{balls of that colour}}{\text{all balls}}
  • One after another, without replacement

    Without replacement

    P(A then B)=an×bn−1P(A \text{ then } B) = \dfrac{a}{n} \times \dfrac{b}{n - 1}
  • With replacement: the bag resets

    j of k draws, with replacement

    P(j of k)=kCj pj(1−p)k−jP(j \text{ of } k) = {}^kC_j\, p^j (1 - p)^{k - j}

Watch out for (5)

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