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  1. Home
  2. 12th grade
  3. Statistics & Probabilities
  4. Exercise : Calculate basic probabilities

Calculate basic probabilities Statistics & Probabilities

Paul throws an unbiased die.

Let A be the event : "the result is even".

Let B be the event : "the result of the die is greater than or equal to 3".

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

As all outcomes are equally likely, the probability of an event E is defined as the number of outcomes favorable to the event E divided by the total number of possible outcomes.

Step 1

Number of possible outcomes

Paul is throwing a unbiased die. The possible outcomes are:

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

Therefore, the total number of possible outcomes is 6.

Step 2

Number of favorable outcomes

Since we want both events A and B to occur, the die must give an even number and greater or equal to 3. The favorable outcomes are:

\left\{ 4, 6 \right\}

Therefore, the number of favorable outcomes is 2.

Step 3

Conclusion

The probability is:

P\left(A\cap B\right)=\dfrac{2}{6}

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

In a bag there are 5 red marbles, 10 blue marbles, 15 green marbles, and 20 yellow marbles. Joey picks a marble.

What is the probability of Joey picking a blue marble?

Let E be the event "Joey pick a blue marble".

As all outcomes are equally likely, the probability of an event E is defined as the number of outcomes favorable to the event E divided by the total number of possible outcomes.

Step 1

Number of possible outcomes

The total number of possible outcomes is the total number of marbles in the bag:

5+10+15+20=50

Step 2

Number of favorable outcomes

Since Joey wants to pick a blue marble, the number of favorable outcomes is the number of blue marbles: 10.

Step 3

Conclusion

The probability is:

P\left(E\right)=\dfrac{10}{50}=\dfrac{1}{5}.

P\left(E\right)=\dfrac{1}{5}

In a bag there are 10 red cards numbered 1 through 10 and 7 blue cards numbered 1 through 7.

Bob is picking a card.

Let A be the event: "the color of the card is red."

Let B be the event: "the number on the card is greater than 6."

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

As all outcomes are equally likely, the probability of an event E is defined as the number of outcomes favorable to the event E divided by the total number of possible outcomes.

Step 1

Number of possible outcomes

The total number of possible outcomes is the total number of cards in the bag:

7+10=17

Step 2

Number of favorable outcomes

Since we want both events A and B to occur, the color of the card must be red and the number on the card must be greater than 6. The favorable outcomes are:

\left\{ Red7, Red8, Red9, Red10\right\}

Therefore, the number of favorable outcomes is 4.

Step 3

Conclusion

The probability is:

P\left(A\cap B\right)=\dfrac{4}{17}

P\left(A\cap B\right)=\dfrac{4}{17}

At a local car dealership there are 20 new trucks and 19 new cars. Also, at the same dealership there are 5 second-hand trucks and 16 second-hand cars.

Determine the probability that a randomly selected vehicle is new.

Let E be the event "The selected vehicle is new".

As all outcomes are equally likely, the probability of an event E is defined as the number of outcomes favorable to the event E divided by the total number of possible outcomes.

Step 1

Number of possible outcomes

The total number of possible outcomes is the total number of vehicles at the local dealership:

20+19+5+16=60

Step 2

Number of favorable outcomes

The number of favorable outcomes is the number of new vehicles:

20+19=39

Step 3

Conclusion

The probability is:

P\left(E\right)=\dfrac{39}{60}=\dfrac{13}{20}.

P\left(E\right)=\dfrac{13}{20}

Mark throws an unbiased die.

Let A be the event : "the result is odd".

Let B be the event : "the result of the die is greater than or equal to 4".

Find P\left(A\cup B\right).

As all outcomes are equally likely, the probability of an event E is defined as the number of outcomes favorable to the event E divided by the total number of possible outcomes.

Step 1

Number of possible outcomes

Mark is throwing a unbiased die. The possible outcomes are:

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

Therefore, the total number of possible outcomes is 6.

Step 2

Number of favorable outcomes

For the event A, the favorable outcomes are:

\left\{ 1, 3, 5 \right\}

In conclusion, the probability of event A is:

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

For the event B, the favorable outcomes are:

\left\{ 4, 5, 6 \right\}

In conclusion, the probability of event B is:

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

For the A\cap B, we want both events A and B to occur, so the dice must give an odd number greater than or equal to 4. The only favorable outcome is:

\left\{ 5 \right\}

Therefore the number of favorable outcomes is 1.

In conclusion, the probability of event A\cap B is:

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

Step 3

Conclusion

The formula to calculate the probability of a union is:

P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)

The final probability is:

P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{5}{6}.

P\left(A\cup B\right)=\dfrac{5}{6}

In a bag there are 12 green cards numbered 1 through 12 and 10 yellow cards numbered 1 through 10.

Luke is picking a card.

Let A be the event: "the color of the card is yellow."

Let B be the event: "the number on the card is greater than 8."

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

Step 1

Number of possible outcomes

As all outcomes are equally likely, the probability of an event E is defined as the number of outcomes favorable to the event E divided by the total number of possible outcomes.

The total number of possible outcomes is the total number of cards in the bag:

12+10=22

Step 2

Number of favorable outcomes

Since we want both events A and B to occur, the color of the card must be yellow and the number on the card must be greater than 8. The favorable outcomes are:

\left\{ Yellow9, Yellow10 \right\}

Therefore, the number of favorable outcomes is 2.

Step 3

Conclusion

The probability is:

P\left(A\cap B\right)=\dfrac{2}{22}=\dfrac{1}{11}.

P\left(A\cap B\right)=\dfrac{1}{11}

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See also
  • Course : Introduction to probabilities
  • Exercise : Determine whether events are dependent or independent
  • Exercise : Use the multiplication rule to determine the probability of an intersection of events
  • Exercise : Calculate conditional probabilities
  • Exercise : Use permutations and combinations to determine the number of outcomes of an experiment
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