Loading [MathJax]/jax/output/CommonHTML/jax.js

hacklink al hack forum organik hit kayseri escort onwin güncel giriş grandpashabet Mostbet 토토사이트 Sakarya escort Sakarya escort Ganobet Giriş sex izle Gaziantep Escort deneme bonusu veren siteler deneme bonusu grandpashabet iptv iptv satın al grandpashabet grandpashabet ücretsiz url kısaltma halkali escort güvenilir bahis siteleri grandpashabet ultrabet ultrabet ultrabet safirbet teslabahis Yorkbet safirbet 1 Betcio jojobet Palacebet holiganbet zbahis jojobet jojobet Türk İfşa Vip türk ifşa ultrabet ultrabet giriş ultrabet ultrabet giriş ultrabet ultrabet twitter casibom giriş İzmit escort padişahbet padişahbet Manavgat escort jojobet tipobet venüsbet Soft2bet

Topics

>

>

>

Probability Tree (Independent Events)

>

Probability Tree (Independent)

Probability Tree (Independent)

Try VividMath Premium to unlock full access

Start Free Trial

Loading...

Lets see your results!

Points

Score

Time

1
2
3
4
5
6

Answered
Review

Question 1 of 6

Find the probability of tossing a normal coin twice and getting:

$(a) 2$ Tails

$(b)$ Heads and Tails (in any order)

Write fractions in the format “a/b”

$(a)$ (1/4)

$(b)$ (1/2, 2/4)

Incorrect

Help Video

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Addition Rule

$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

$(a)$ Find the probability of getting

$2$ Tails.

First, set up a probability tree showing all possible outcomes of tossing a coin twice

Find the probability of all outcomes

$\mathsf{P(Heads)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ Heads out of $2$ sides

$\mathsf{P(Tails)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ Tails out of $2$ sides

Mark each outcome with its corresponding probability

Finally, solve for the probability by picking the branch that leads to

$2$ Tails and multiplying the probabilities along it

$1/2$

$times$ $1/2$

$=$

$1/4$

Product Rule

Therefore, the probability of throwing Tails twice is

$1/4$

$(b)$ Find the probability of getting Heads and Tails.

Use the probability tree from part

$(a)$ and pick the branches that lead to Heads and Tails then multiply the probabilities along it

First branch (HT):

$1/2$

$times$ $1/2$

$=$

$1/4$

Product Rule

Second branch (TH):

$1/2$

$times$ $1/2$

$=$

$1/4$

Product Rule

Finally, add the solved probability for each branch

$1/4$

$+$ $1/4$

$=$

$2/4$

$=$

$1/2$

Addition Rule

Therefore, the probability of getting Heads and Tails is

$1/2$

$(a) 1/4$

$(b) 1/2$

Question 2 of 6

Find the probability of tossing a normal coin thrice and getting:

$(a)$ Heads, Tails, Tails (in that order)

$(b) 2$ Tails and Heads (in any order)

Write fractions in the format “a/b”

$(a)$ (1/8)

$(b)$ (3/8)

Incorrect

Help Video

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Addition Rule

$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

$(a)$ Find the probability of getting Heads, Tails, Tails.

First, set up a probability tree showing all possible outcomes of tossing a coin thrice

Find the probability of all outcomes

$\mathsf{P(Heads)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ Heads out of $2$ sides

$\mathsf{P(Tails)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ Tails out of $2$ sides

Mark each outcome with its corresponding probability

Finally, solve for the probability by picking the branch that leads to Heads, Tails, Tails and multiplying the probabilities along it

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Therefore, the probability of throwing Heads, Tails, Tails twice in that order is

$1/8$

$(b)$ Find the probability of getting

$2$ Tails and

$1$ Heads.

Use the probability tree from part

$(a)$ and pick the branches that lead to

$2$ Tails and

$1$ Heads then multiply the probabilities along it

First branch (HTT):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Second branch (THT):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Third branch (TTH):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Finally, add the solved probability for each branch

$1/8$

$+$ $1/8$

$+$ $1/8$

$=$

$3/8$

Addition Rule

Therefore, the probability of getting

$2$ Tails and

$1$ Heads is

$3/8$

$(a) 1/8$

$(b) 3/8$

Question 3 of 6

Find the probability of tossing a normal coin thrice and getting:

$(a) 3$ of a kind

$(b)$ at least

$2$ Heads

Write fractions in the format “a/b”

$(a)$ (1/4, 2/8)

$(b)$ (1/2, 4/8)

Incorrect

Help Video

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Addition Rule

$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

$(a)$ Find the probability of getting

$3$ of a kind.

First, set up a probability tree showing all possible outcomes of tossing a coin thrice

Find the probability of all outcomes

$\mathsf{P(Heads)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ Heads out of $2$ sides

$\mathsf{P(Tails)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ Tails out of $2$ sides

Mark each outcome with its corresponding probability

Pick the branches that lead to

$3$ of a kind then multiply the probabilities along it

First branch (HHH):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Second branch (TTT):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Finally, add the solved probability for each branch

$1/8$

$+$ $1/8$

$=$

$2/8$

$=$

$1/4$

Addition Rule

Therefore, the probability of getting

$3$ of a kind is

$1/4$

$(b)$ Find the probability of getting at least

$2$ Heads.

Use the probability tree from part

$(a)$ and pick the branches that lead to at least

$2$ Heads then multiply the probabilities along it

First branch (HHT):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Second branch (HTH):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Third branch (HHT):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Fourth branch (HHH):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Finally, add the solved probability for each branch

$1/8$

$+$ $1/8$

$+$ $1/8$

$+$ $1/8$

$=$

$4/8$

$=$

$1/2$

Addition Rule

Therefore, the probability of getting at least

$2$ Heads is

$1/2$

$(a) 1/4$

$(b) 1/2$

Question 4 of 6

Find the probability of tossing a normal coin thrice and getting

$3$ Heads or Heads, Tails, Heads in that order.

Write fractions in the format “a/b”

(1/4, 2/8)

Incorrect

Help Video

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Addition Rule

$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

First, set up a probability tree showing all possible outcomes of tossing a coin thrice

Find the probability of all outcomes

$\mathsf{P(Heads)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ Heads out of $2$ sides

$\mathsf{P(Tails)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ Tails out of $2$ sides

Mark each outcome with its corresponding probability

Pick the branches that lead to the preferred outcomes then multiply the probabilities along it

First branch (HHH):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Second branch (HTH):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Finally, add the solved probability for each branch

$1/8$

$+$ $1/8$

$=$

$2/8$

$=$

$1/4$

Addition Rule

Therefore, the probability of getting

$3$ Heads and Heads, Tails, Heads is

$1/4$

$1/4$

Question 5 of 6

A couple plan to have three kids. Find the probability of having:

$(a) 2$ Girls and

$1$ Boy (in any order)

$(b) 3$ of the same gender

Write fractions in the format “a/b”

$(a)$ (3/8)

$(b)$ (1/4, 2/8)

Incorrect

Help Video

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Addition Rule

$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

$(a)$ Find the probability of having

$2$ Girls and

$1$ Boy.

First, set up a probability tree showing all possible genders of

$3$ kids

Find the probability of all outcomes

$\mathsf{P(Girl)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ out of $2$ possible genders

$\mathsf{P(Boy)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$	$1$ out of $2$ possible genders

Mark each outcome with its corresponding probability

Pick the branches that lead to

$2$ Girls and

$1$ Boy

First branch (BGG):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Second branch (GBG):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Third branch (GGB):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Finally, add the solved probability for each branch

$1/8$

$+$ $1/8$

$+$ $1/8$

$=$

$3/8$

Addition Rule

Therefore, the probability of having

$2$ Girls and

$1$ Boy is

$3/8$

$(b)$ Find the probability of having

$3$ of the same sex.

Use the probability tree from part

$(a)$ and pick the branches that lead to

$3$ of the same sex then multiply the probabilities along it

First branch (BBB):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Second branch (GGG):

$1/2$

$times$ $1/2$

$times$ $1/2$

$=$

$1/8$

Product Rule

Finally, add the solved probability for each branch

$1/8$

$+$ $1/8$

$=$

$2/8$

$=$

$1/4$

Addition Rule

Therefore, the probability of having

$3$ of the same sex is

$1/4$

$(a) 3/8$

$(b) 1/4$

Question 6 of 6

A box contains

$5$ balls,

$3$ of which are Yellow and

$2$ are Blue. Find the probability of drawing

$3$ balls at random and getting:

$(a) 3$ Yellow

$(b)$ at least

$1$ Blue

Write fractions in the format “a/b”

$(a)$ (27/125)

$(b)$ (98/125)

Incorrect

Help Video

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Complementary
Probability

$\mathsf{P(\dot{E})}=1-\mathsf{P(E)}$

Addition Rule

$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

$(a)$ Find the probability of drawing

$3$ Yellow balls.

First, set up a probability tree showing all possible outcomes of drawing

$3$ balls

Find the probability of all outcomes

$\mathsf{P(Yellow)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{3}}{\color{#007DDC}{5}}$	$3$ Yellow balls out of $5$ balls

$\mathsf{P(Blue)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$	$2$ Blue balls out of $5$ balls

Mark each outcome with its corresponding probability

Finally, pick the branch that lead to

$3$ Yellow and multiply the probabilities along it

$3/5$

$times$ $3/5$

$times$ $3/5$

$=$

$27/125$

Product Rule

Therefore, the probability of drawing at

$3$ Yellow balls is

$27/125$

$(b)$ Find the probability of drawing at least

$1$ Blue ball.

Notice that all outcomes have at least

$1$ Blue except YYY

Hence, we can simply get the complement of P(YYY)

$\mathsf{P(\dot{YYY})}$	$=$	$1-\mathsf{P(YYY)}$	Complementary Probability

	$=$	$1-27/125$	Substitute value

	$=$	$125/125-27/125$

	$=$	$98/125$

Therefore, the probability of drawing at least

$1$ Blue is

$98/125$

$(a) 27/125$

$(b) 98/125$