Practice: multi-stage experiments

Probability, grade 11 (5 units) Lesson 11 of 16Practice

Practice: multi-stage experiments

1

In an entrance exam to a college 20% of the candidates were from kibbutzim, 40% from moshavim and 40% from towns.
70% of the candidates passed the exam.
$\displaystyle \frac{1}{8}$ of the candidates from moshavim failed the exam.
The probability of choosing at random a candidate who was from a town and passed the exam is 2.5 times the probability of choosing at random a candidate who was from a kibbutz and passed the exam.

a

Write $\displaystyle x$ for the probability of choosing at random a candidate who was from a kibbutz and passed the exam. Write an equation and find $\displaystyle x$ .

b

Complete the probabilities in the tree: the first stage – where the candidate is from, the second stage – passed or failed.

c

A candidate is chosen at random from those who failed the exam. What is the probability that the candidate was not from a town?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

d

Moshe passed the exam. What is the probability that he was not from a moshav?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

e

Five candidates are chosen at random from those who passed the exam. What is the probability that at least one of them was from a moshav?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

2

A set of different digits is given: 3 even digits (not 0), and the rest odd digits.
Yoni makes a two-digit number from the digits in the set.
The first digit he chooses at random is the tens digit, and the second digit he chooses at random is the units digit.
Yoni chooses each digit at most once – without putting back.

a

The probability that Yoni makes an odd number is $\displaystyle \frac{4}{7}$ . How many odd digits are in the set? Write $\displaystyle x$ for the number of odd digits, and solve an equation.

b

The number Yoni made is known to be even. What is the probability that both digits he chose are even? Complete the tree, then work it out.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

Emily makes a three-digit number from the same set, also without putting back: the hundreds digit, then the tens digit, then the units digit.
The first digit Emily chose is known to be even.
What is the probability that the sum of the digits of her number is even? In the tree – the two digits she chooses after the first one.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

3

A tennis player plays three sets.
The probability that she wins the first set is $\displaystyle 0.6$ .
After a set she won, the probability that she wins the next set is $\displaystyle 0.7$ .
After a set she lost, the probability that she wins the next set is $\displaystyle 0.4$ .
Complete the probabilities in the tree.

a

What is the probability that she wins at least two of the three sets?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

She is known to have won at least two sets. What is the probability that she lost the first set? Round to three decimal places.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

4

A test has three multiple-choice questions, each with four answers, only one of them right.
Tamar did not study, and chooses an answer at random in each question.
Complete the probabilities in the tree.

a

What is the probability that Tamar answers exactly one question right?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

What is the probability that Tamar answers at least one question right?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

5

Bag I holds two red balls and three blue ones.
Bag II holds three red balls and two blue ones.
A coin is tossed: on heads bag I is chosen, on tails – bag II.
Two balls are taken at random from the chosen bag, one after the other, without putting back.
Complete the probabilities in the tree.

a

What is the probability that both balls are red?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

Both balls are known to be red. What is the probability that they were taken from bag I?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

Both balls are known to be the same colour. What is the probability that they were taken from bag I?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: probability, grade 11 (5 units)

The rules of probability

Equally likely outcomes – $\displaystyle k$ the outcomes in the event $\displaystyle A$ , $\displaystyle n$ the outcomes in the sample space
$\displaystyle P(A)=\frac{k}{n}$
Every event; $\displaystyle {P(\emptyset)}={0}$ , and $\displaystyle {P(U)}={1}$ for the sample space $\displaystyle U$
$\displaystyle 0\le P(A)\le1$
An event contained in another event
$\displaystyle P(A)\le P(B)$ $\displaystyle A\subseteq B$
The complement $\displaystyle \overline{A}$ – $\displaystyle A$ does not happen
$\displaystyle P(\overline{A})$ $\displaystyle =$ $\displaystyle 1-P(A)$
The union – $\displaystyle A$ or $\displaystyle B$ (or both); $\displaystyle A\cap B$ – the intersection, both $\displaystyle A$ and $\displaystyle B$
$\displaystyle P(A\cup B)$ $\displaystyle =$ $\displaystyle P(A)+P(B)-P(A\cap B)$
Disjoint events – no two of them have an outcome in common
$\displaystyle P(A_1\cup A_2\cup\dots\cup A_n)=P(A_1)+P(A_2)+\dots+P(A_n)$

Conditional probability and independence

The probability of $\displaystyle A$ given $\displaystyle B$
$\displaystyle P(A/B)=\frac{P(A\cap B)}{P(B)}$
A path in a tree – the product of the probabilities on its branches
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(B)\cdot P(A/B)$
Independent events
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(A)\cdot P(B)$
Independent events – knowing that $\displaystyle B$ happened does not change the probability of $\displaystyle A$
$\displaystyle P(A/B)$ $\displaystyle =$ $\displaystyle P(A)$
The law of total probability – the paths that end in $\displaystyle A$ , added up
$\displaystyle P(A)=P(B)\cdot P(A/B)+P(\overline{B})\cdot P(A/\overline{B})$

Counting and Bernoulli's formula

$\displaystyle n$ factorial; $\displaystyle {0!}={1}$
$\displaystyle n!=1\cdot2\cdot3\cdot\dots\cdot n$
Ordered selections – $\displaystyle k$ different objects out of $\displaystyle n$ , one after another
$\displaystyle n\cdot(n-1)\cdot\dots\cdot(n-k+1)=\frac{n!}{(n-k)!}$
The binomial coefficient – the ways to choose $\displaystyle k$ objects out of $\displaystyle n$ , order not counted
$\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}$
Bernoulli’s formula – the probability of $\displaystyle k$ successes in $\displaystyle n$ independent trials, with the probability of success $\displaystyle p$ in each
$\displaystyle P_n(k)=\binom{n}{k}p^k(1-p)^{n-k}$
At least one success in $\displaystyle n$ trials – by the complement
$\displaystyle 1-P_n(0)=1-(1-p)^n$