Mastering Independent Events through Multiplication Rule and Sample Space

	Favorable Outcomes	Total Outcomes	Substitute
$P (Q \cap H)$	$1$	$8$	$\frac{1}{8}$
$P (Q)$	$2$	$8$	$\frac{2}{8}$
$P (H)$	$2$	$8$	$\frac{2}{8}$

Favorable Outcomes

Total Outcomes

Substitute

P (Q \cap H)

1

8

\frac{1}{8}

P (Q)

2

8

\frac{2}{8}

P (H)

2

8

\frac{2}{8}

Number of Girls	Outcome
$111222$	$G_{1} B G_{2} B G_{3} B G_{1} G_{2} G_{1} G_{3} G_{2} G_{3}$	$B G_{1} B G_{2} B G_{3} G_{2} G_{1} G_{3} G_{1} G_{3} G_{2}$

Number of Girls

Outcome

111222

G_{1} B G_{2} B G_{3} B G_{1} G_{2} G_{1} G_{3} G_{2} G_{3}

B G_{1} B G_{2} B G_{3} G_{2} G_{1} G_{3} G_{1} G_{3} G_{2}

On her way to school, Tiffaniqua passes two crossings with red lights. The following tree diagram shows the probabilities of encountering a red light or a green light at the two crossings.

a tree diagram showing the sample space of encountering red lights

Calculate the following probabilities in percent.

Tiffaniqua encounters at most one red light.

Tiffaniqua encounters two green lights.

The probability of Tiffaniqua encountering at most one red light means she can encounter one or no red lights. We have marked these below.

To find this, we can determine the probability that Tiffaniqua encounters two red lights and then calculate the complement of this. To calculate the probability of encountering two red lights, we use the Multiplication Rule of Probability. P(R,R)&=0.6* 0.6 &= 0.36 Now, we determine the complement of this by subtracting P(R,R) from 1. 1- 0.36=0.64 The probability of encountering at most one red light is 64 %.

As in Part A, we will use the Multiplication Rule of Probability to determine the probability of encountering two green lights. P(G,G)&=0.4* 0.4 &= 0.16

During a game of basketball, Shaquille O'Neal gets three penalty shots. The statistics show that during earlier games he managed to make $67$ out of $108$ shots.

What is the probability of Shaq making all three penalty shots? Round your answer to the nearest percent.

What is the probability that Shaq makes at least one of the penalty shots? Round your answer to the nearest percent.

In this case we are working with experimental probabilities. We know that Shaq historically has made 67 out of 108 shots. Therefore, the probability of Shaq making any given shot is 67108. Let's draw a tree diagram and mark the path showing Shaq making all of the shots.

Using the Multiplication Rule of Probability, we can determine the probability of Shaq making all three shots.

There is a 24 % probability of Shaq making all three shots.

The probability of Shaq making at least one shot is the complement of Shaq making none of the shots. From the exercise we know that the probability of making one shot is 67108. This means the probability of Shaq not making a shot is 1 minus 67108.

Now we can use the Multiplication Rule of Probability to determine the probability of Shaq not making any of the shots.

Now we can determine the complement of this. P(at least one shot)&=1- 0.05 &=0.95 There is a 95 % probability of Shaq making at least one shot.

Diego is flipping a coin. What is the probability of Diego getting three tails in a row? Answer with a fraction in its simplest form.

If we flip a coin there are two outcomes: heads and tails. Since both outcomes are equally probable, we know that the probability of getting tails on one flip is one-half. Notice that separate coin flips are independent of each other. Let's mark the outcome and related probability of tossing three tails in a tree-diagram.

According to the Multiplication Rule of Probability, we determine the probability of multiple events happening by multiplying their individual probabilities.

The probability of getting three tails is 18.

Dylan is practicing throwing darts. He has worked out that the probability of hitting the target on a single throw is on average $0.8 .$ He throws two darts at the target. What is the average probability that both throws are a hit? Answer in percent.

Notice that each throw is independent of each other. Therefore, the average probability of hitting the target is 80 % for every throw. Let's illustrate this in a tree diagram.

To obtain the probability of multiple events occurring, we will use the Multiplication Rule of Probability. This means we must multiply their individual probabilities.

Heichi is choosing beads from a jar that contains $40$ red beads and $60$ orange beads.

What is the probability of Heichi randomly picking two red beads if the first bead is replaced? Answer in percent.

What is the probability of Heichi randomly picking two red beads if the first bead is not replaced? Answer in percent and round to one decimal.

If the first red bead is replaced, we have a probability of 40100 for picking a red bead the second time. Let's show this in a tree diagram.

Using the Multiplication Rule of Probability, we can determine the probability of picking two red beads.

If the first red bead is not replaced, the probability of picking the second bead changes. This is because the number of red beads and the total number of beads is one less after the first one is picked.

Again, to determine the probability of picking two red beads in a row, we will use the Multiplication Rule of Probability.

Which of the following options show independent events?

i ii iii iv P (A \cap B) = \frac{1}{4}, P (A) = \frac{3}{4}, P (B) = \frac{1}{4} P (A \cap B) = \frac{2}{9}, P (A) = \frac{2}{3}, P (B) = \frac{1}{3} P (A \cap B) = \frac{1}{3}, P (A) = \frac{1}{2}, P (B) = \frac{2}{3} P (A \cap B) = \frac{4}{1 5}, P (A) = \frac{2}{6}, P (B) = \frac{4}{6}

The expression P(A ⋂ B) means that A and B both occur. If these are independent events, the product of the individual probabilities — P(A) and P(B) — equals P(A ⋂ B). P(A ⋂ B)= P(A)* P(B) Let's try this for each of the probabilities. i:& 1/4? = 3/4* 1/4 &&⇒ 1/4 ≠ 3/16 &&& -10pt * [1em] ii:& 2/9? = 2/3* 1/3 &&⇒ 2/9 = 2/9 &&& -10pt ✓ [1em] iii:& 1/3? = 1/2* 2/3 &&⇒ 1/3 = 1/3 &&& -10pt ✓ [1em] iv:& 4/15? = 2/6* 4/6 &&⇒ 4/15 ≠ 2/9 &&& -10pt * [1em] As we can see, ii and iii show independent events.

According to Mega Millions website, the odds of winning the jackpot are about

1

out of

300000000 .

And, as reported by the Center for Disease Control and Prevention website, the chances of being struck by lightning in a given year are around

1

500000 .

With this information in mind, which of the following options are more likely to happen than winning the Mega Millions jackpot?

A B C D Getting struck by lightning two consecutive years Getting heads in a coin toss 25 consecutive times Both of the above None of the above

We will calculate the probabilities of the both events provided, one at a time.

Being Struck by a Lightning Two Consecutive Years

It is important to notice that the probability of being struck by a lightning at a given year P(lightning) is independent on whether or not one was struck by a lightning the previous year. Because these are independent events, the probability of being struck by a lightning two years consecutively can be calculated by using the Multiplication Rule of Probability.

Since 1250 000 000 000 < 1300 000 000, it can be concluded that the probability of getting hit by a lightning two consecutive years is less likely than winning the Mega Millions jackpot.

Getting Heads 25 Consecutive Times

Once more, because the result between each coin toss is independent, the Multiplication Rule of Probability can be used.

Since 133 554 432 > 1300 000 000, getting heads 25 times in a row is more likely than winning the Mega Millions jackpot.

Probability and the Independence of Events

Catch-Up and Review

Selecting Two Students From a Group of Four Students

Investigating the Probability of Drawing a Marble

Independent and Dependent Events

Independent Events

Why

Dependent Events

Why

Finding the Probability of Two Events Occurring at the Same Time Using the Multiplication Rule of Probability

Hint

Solution

Rolling a Die and Deciding Whether the Given Events Are Independent

Hint

Solution

Identifying Dependent and Independent Events

Deciding Whether the Given Events About Euro $2020$ Predictions Are Independent

Hint

Solution

Deciding Whether the Given Events When Picking a Card Are Independent

Hint

Solution

Calculating Probability of an Event Using Independency of Events

Deciding Whether the Events Are Independent or Dependent

Hint

Solution

Being Struck by a Lightning Two Consecutive Years

Getting Heads 25 Consecutive Times

Probability and the Independence of Events

Recommended exercises

	11 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions