Probability Formulas

Question 1:

In a bag, there are 7 red marbles, 5 blue marbles, and 4 green marbles. Two marbles are randomly selected from the bag without replacement. What is the probability that both of them are blue?

Solution:

For the first draw, Probability of selecting a blue marble is 5/16 (5 blue marbles out of 16 total marbles).

After the first marble is drawn, there are 4 blue marbles left out of 15 total marbles.

For the second draw, the probability of selecting a blue marble is 4/15.

To calculate the probability of both marbles being blue => Probability = \frac{5}{16}\times \frac{4}{15}

Probability = \frac{20}{24} = \frac{1}{12}

Question 2:

In a drawer, there are 4 black pens, 3 blue pens, and 5 red pens. A pen is drawn at random from the drawer. What is the probability that it is either black or blue?

Solution:

We have calculate the probability of drawing a black or blue pen from the drawer

The total no. of pens in the drawer is 4 black + 3 blue + 5 red = 12 pens

Probability of drawing a black pen is 4/12

Probability of drawing a blue pen is 3/12 = 1/4

Hence, The probability of drawing either a black or blue pen, we add the individual probabilities:

Probability = \frac{4}{12} + \frac{3}{12}
Probability = \frac{7}{12}

Question 3:

In a bag, there are 3 green bulbs, 4 orange bulbs, and 5 white bulbs. A bulb is randomly picked from the bag. What is the probability of selecting either a green bulb or a white bulb?

Solution:

Total number of bulbs in the bag is 3 green + 4 orange + 5 white = 12 bulbs

Number of green bulbs is 3, and the number of white bulbs is 5

Probability of selecting either a green or a white bulb, we add the number of green bulbs and the number of white bulbs, and then divide it by the total number of bulbs.

Probability = (Number of green bulbs + Number of white bulbs) / Total number of bulbs
Probability = \frac{3 + 5}{12}
Probability = \frac{8}{12}
Probability = \frac{2}{3}

Question 4:

Sylvester Stallone brought a box of balloons for a group of students. The box contains 3 balloons of Shape A, 4 balloons of Shape B, and 5 balloons of Shape C. If three balloons are randomly drawn from the box, what is the probability that all three balloons are of different shapes?

Solution:

Total No. of Balloons = 3 Balloons of Shape A +4 Balloons of Shape B + 5 Balloons of Shape C = 12

n(s)= ^{12}C_{3}=220

n(e)= ^{3}C_{1} *  ^{4}C_{1}*  ^{5}C_{1} =60

P= \frac{60}{220}=\frac{3}{11}

Question 5:

Joey Tribbiani organized a rack race with two participants. The probability of the first participant winning is 2/7, and the probability of the second participant winning is 3/5. What is the probability that one of them will win?

Answers: 

Let’s denote:
P(A) = Probability of the first participant winning = 2/7
P(B) = Probability of the second participant winning = 3/5

The probability of both participants winning simultaneously (a tie) is zero since there can only be one winner. Therefore, the probability that one of them will win is:

P(one of them wins) = P(A) + P(B) – P(A and B)
P(one of them wins) = P(A) + P(B) – 0 (since P(A and B) = 0)
P(one of them wins) = P(A) + P(B)

Substituting the given probabilities:
P(one of them wins) = \frac{2}{7}+\frac{3}{5}
P(one of them wins) =\frac{10}{35}+\frac{21}{35}
P(one of them wins) = \frac{31}{35}