Probability — Practice

Question 1

A bag contains 5 red and 3 blue balls. Two balls are drawn one after another without replacement. Find the probability that both are red.

Accepted Answer

Let A = first ball is red, B = second ball is red. P(A) = 5/8. P(B|A) = 4/7 (4 red balls remain from 7 total after one red drawn). P(A∩B) = P(A)·P(B|A) = (5/8)·(4/7) = 20/56 = 5/14.

Question 2

There are two bags. Bag I: 3 red, 4 black balls. Bag II: 5 red, 6 black balls. One bag is selected at random (equal probability) and one ball is drawn. The ball is red. Find the probability that it came from Bag I.

Accepted Answer

Let E₁ = Bag I selected, E₂ = Bag II selected, A = red ball drawn. P(E₁) = P(E₂) = 1/2. P(A|E₁) = 3/7. P(A|E₂) = 5/11. TOTAL PROBABILITY: P(A) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) = (1/2)(3/7) + (1/2)(5/11) = 3/14 + 5/22 = 33/154 + 35/154 = 68/154 = 34/77. BAYES' THEOREM: P(E₁|A) = P(E₁)P(A|E₁) / P(A) = (1/2)(3/7) / (34/77) = (3/14) / (34/77) = (3/14) × (77/34) = 231/476 = 33/68.

Question 3

The probability that a student passes an exam is 2/3. In a group of 5 students, find: (i) P(at least 2 pass), (ii) the mean and variance of the number of students who pass.

Accepted Answer

X = number of students who pass ~ B(n=5, p=2/3, q=1/3). P(X=r) = ⁵Cᵣ·(2/3)ʳ·(1/3)^(5−r). (i) P(X ≥ 2) = 1 − P(X=0) − P(X=1). P(X=0) = ⁵C₀·(2/3)⁰·(1/3)⁵ = 1·1·1/243 = 1/243. P(X=1) = ⁵C₁·(2/3)¹·(1/3)⁴ = 5·(2/3)·(1/81) = 10/243. P(X ≥ 2) = 1 − 1/243 − 10/243 = 1 − 11/243 = 232/243. (ii) MEAN: E(X) = np = 5·(2/3) = 10/3. VARIANCE: Var(X) = npq = 5·(2/3)·(1/3) = 10/9. (Standard deviation = √(10/9) = √10/3.)

Practice — Probability

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities