Use the multiplication rule to determine the probability of an intersection of events - High school Statistics & Probabilities Exercises

James throws two unbiased dice.

Let A be the event : "the result of the first die is odd".
Let B be the event : "the result of the second die is 4 or less".

Calculate P\left(A\cap B\right).

P\left(A\cap B\right)=0.4

P\left(A\cap B\right)=\dfrac13

P\left(A\cap B\right)=\dfrac23

P\left(A\cap B\right)=\dfrac56

We can see that the probability that event A occurs does not affect the probability of event B occurring. The events are therefore independent and we have:

P\left(A \cap B\right) = P\left(A\right).P\left(B\right)

If an event E can happen in r ways and n is the size of the sample space, then the probability of the occurrence of E is:

p\left(E\right)=\dfrac{r}{n}

Here, we have n=6 as the sample space is the set:

\{1{,}2{,}3{,}4{,}5{,}6\}

Event A happens if one of the numbers 1{,}3 or 5 appears. So:

P\left(A\right) = \dfrac{|\{1{,}3{,}5\}|}{6}=\dfrac{3}{6} = \dfrac{1}{2}

Event B happens if one of the numbers 1{,}2{,}3 or 4 appears. Thus:

P\left(B\right) = \dfrac{|\{1{,}2{,}3{,}4\}|}{6} = \dfrac{4}{6} = \dfrac{2}{3}

Therefore, we have:

P\left(A \cap B\right) = \dfrac{1}{2} \times \dfrac{2}{3}= \dfrac{1}{3}

P\left(A\cap B\right)=\dfrac{1}{3}

In a standard 52 card deck, we draw two cards one by one and replace them each time.

Let A be the event : "the first card is a queen".
Let B be the event : "the second card is a heart".

Calculate P\left(A\cap B\right).

P\left(A\cap B\right)=\dfrac16

P\left(A\cap B\right)=\dfrac1{13}

P\left(A\cap B\right)=0.1

P\left(A\cap B\right)=\dfrac1{52}

We can see that the probability that event A occurs does not affect the probability of event B occurring. The events are therefore independent and we have:

P\left(A \cap B\right) = P\left(A\right).P\left(B\right)

If an event E can happen in r ways and n is the size of the sample space, then the probability of the occurrence of E is:

p\left(E\right)=\dfrac{r}{n}

Here, we have n=52.

Since we have four queens in a deck:

P\left(A\right) = \dfrac{4}{52}=\dfrac{1}{13}

Since we have 13 hearts in a deck:

P\left(B\right) = \dfrac{13}{52} = \dfrac{1}{4}

Therefore, we have:

P\left(A \cap B\right) = \dfrac{1}{4} \times \dfrac{1}{13}= \dfrac{1}{52}

P\left(A\cap B\right)=\dfrac{1}{52}

In a family with three children:

Let A be the event : "the first two children are girls".
Let B be the event : "the third child is a boy".

Calculate P\left(A\cap B\right).

P\left(A\cap B\right)=0.2

P\left(A\cap B\right)=\dfrac{1}{4}

P\left(A\cap B\right)=\dfrac{1}{6}

P\left(A\cap B\right)=\dfrac{1}{8}

We can see that the probability that event A occurs does not affect the probability of event B occurring. The events are therefore independent and we have:

P\left(A \cap B\right) = P\left(A\right).P\left(B\right)

If an event E can happen in r ways and n is the size of the sample space, then the probability of the occurrence of E is:

p\left(E\right)=\dfrac{r}{n}

For event A, we have n=4 and r=1 since we have 4 possible cases:

\textcolor{Blue}{\left(g,g\right)}, \left(g,b\right), \left(b,g\right),\left(b,b\right)

Thus:

p\left(A\right)=\dfrac{1}{4}

Since the third child is either a boy or a girl, we also have:

p\left(B\right)=\dfrac{1}{2}

Therefore, we have:

P\left(A \cap B\right) = \dfrac{1}{2} \times \dfrac{1}{4}= \dfrac{1}{8}

P\left(A\cap B\right)=\dfrac{1}{8}

A jar contains 7 blue marbles, 4 red marbles, and 9 white marbles. We pick two marble one by one and replace them every time.

Let A be the event : "the first marble is red".
Let B be the event : "the second marble is not white."

Calculate P\left(A\cap B\right).

P\left(A\cap B\right)=0.24

P\left(A\cap B\right)=0.09

P\left(A\cap B\right)=0.11

P\left(A\cap B\right)=0.15

We can see that the probability that event A occurs does not affect the probability of event B occurring. The events are therefore independent and we have:

P\left(A \cap B\right) = P\left(A\right).P\left(B\right)

If an event E can happen in r ways and n is the size of the sample space, then the probability of the occurrence of E is:

p\left(E\right)=\dfrac{r}{n}

Here, we have n=20 as we have 20 marbles in total.

For event A, we have r=4 since we have 4 red marbles:

P\left(A\right) = \dfrac{4}{20}=0.2

For event B, we have r=11 since we have 11 non-white marbles:

P\left(B\right) = \dfrac{11}{20}=0.55

Therefore, we have:

P\left(A \cap B\right) = 0.55 \times 0.2 = 0.11

P\left(A\cap B\right)=0.11

We flip a coin four times:

Let A be the event : "we get exactly two tails on the first tree flips".
Let B be the event : "we get a head on the fourth flip".

Calculate P\left(A\cap B\right).

P\left(A\cap B\right)=\dfrac5{16}

P\left(A\cap B\right)=\dfrac{3}{16}

P\left(A\cap B\right)=\dfrac{2}{9}

P\left(A\cap B\right)=0.4

We can see that the probability that event A occurs does not affect the probability of event B occurring. The events are therefore independent and we have:

P\left(A \cap B\right) = P\left(A\right).P\left(B\right)

If an event E can happen in r ways and n is the size of the sample space, then the probability of the occurrence of E is:

p\left(E\right)=\dfrac{r}{n}

For event A, we have:

\left(h,h,h\right),\left(h,h,t\right),\left(h,t,h\right),\textcolor{Blue}{\left(h,t,t\right)},{\left(t,t,t\right)},\textcolor{Blue}{\left(t,t,h\right)},\textcolor{Blue}{\left(t,h,t\right)},\left(t,h,h\right)

Therefore:

P\left(A\right) = \dfrac{3}{8}

Since either a head or a tail comes up on the last flip, we also have:

P\left(B\right)=\dfrac{1}{2}

Therefore, we have:

P\left(A \cap B\right) =\dfrac{3}{8}\times\dfrac{1}{2}=\dfrac{3}{16}

P\left(A\cap B\right)=\dfrac{3}{16}

We roll a fair die four times:

Let A be the event : "the sum of the first two numbers is 9."
Let B be the event :"the sum of the third and fourth number is 3".

Calculate P\left(A\cap B\right).

P\left(A\cap B\right)=\dfrac{1}{36}

P\left(A\cap B\right)=\dfrac1{216}

P\left(A \cap B\right) =\dfrac{1}{162}

P\left(A\cap B\right)=\dfrac1{72}

We can see that the probability that event A occurs does not affect the probability of event B occurring. The events are therefore independent and we have:

P\left(A \cap B\right) = P\left(A\right).P\left(B\right)

If an event E can happen in r ways and n is the size of the sample space, then the probability of the occurrence of E is:

p\left(E\right)=\dfrac{r}{n}

For both events we have n=36

Event A happens if one of the following cases occur:

\left(3{,}6\right),\left(4{,}5\right),\left(5{,}4\right),\left(6{,}3\right)

Thus:

P\left(A\right) = \dfrac{4}{36}=\dfrac{1}{9}

Event B happens if one of the following cases occur:

\left(1{,}2\right), \left(2{,}1\right)

Therefore:

P\left(A\right) = \dfrac{2}{36}=\dfrac{1}{18}

Thus we have:

P\left(A \cap B\right) = \dfrac{1}{18} \times \dfrac{1}{9}= \dfrac{1}{162}

P\left(A \cap B\right) =\dfrac{1}{162}

In a standard 52 card deck, we draw two cards one by one and replace them every time.

Let A be the event : "the first card is an ace".
Let B be the event : "the second card is a face".

Calculate P\left(A\cap B\right).

P\left(A\cap B\right)=\dfrac{3}{169}

P\left(A\cap B\right)=\dfrac{3}{52}

P\left(A\cap B\right)=0.3

P\left(A\cap B\right)=\dfrac9{26}

We can see that the probability that event A occurs does not affect the probability of event B occurring. The events are therefore independent and we have:

P\left(A \cap B\right) = P\left(A\right).P\left(B\right)

If an event E can happen in r ways and n is the size of the sample space, then the probability of the occurrence of E is:

p\left(E\right)=\dfrac{r}{n}

Here, we have n=52.

The event A happens if we pull out one of the four aces. So:

P\left(A\right) = \dfrac{4}{52}=\dfrac{1}{13}

Event B happens if we pull out a face. We know that there are 3 faces in each suit and that we have 12 faces in total. Thus:

P\left(B\right) = \dfrac{12}{52} = \dfrac{3}{13}

Therefore, we have:

P\left(A \cap B\right) = \dfrac{1}{13} \times \dfrac{3}{13}= \dfrac{3}{169}

P\left(A\cap B\right)=\dfrac{3}{169}