Basic probability formulas and questions

Basic probability formulas

• Probability of an Event A: $P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{n(A)}{n(S)}$
• Complementary Probability: P(A) + P(A') = 1 or, P(A') = 1 - P(A)
• Probability of an impossible event is 0 and of a sure event is 1.
• The number of outcomes(m) favourable to an event cannot be larger than the total number of outcomes
• The probability of occurrence of an event lies in the interval [0,1].

Practice Questions

Question 1. A bag contains 5 white, 7 red and 3 black balls. If 3 balls are drawn one by one without replacement, find the probability that none is red.

Solution:

To find the probability that none of the three balls drawn from the bag are red, we can use basic probability concepts. Here's a step-by-step solution:

Total number of balls: There are 5 white, 7 red, and 3 black balls. So, the total number of balls is: $5 + 7 + 3 = 15$

Total number of ways to draw 3 balls from 15 without replacement: The number of ways to draw 3 balls from 15 is given by the combination formula $^nC_K$, where $n$ is the total number of items, and $k$ is the number of items to choose. So,

$^{15}C_3 = \frac{15!}{3!(15-3)!} = 455$

Number of non-red balls: There are 5 white and 3 black balls, so the total number of non-red balls is: $5 + 3 = 8$

Number of ways to draw 3 non-red balls from these 8: Again using the combination formula, we get: $^8C_3 = \frac{8!}{3!(8-3)!} = 56$

Probability of drawing 3 non-red balls: The probability is the ratio of the number of favorable outcomes to the total number of outcomes. Thus,

$P(\text{none is red}) = \frac{\text{Number of ways to draw 3 non-red balls}}{\text{Total number of ways to draw 3 balls from 15}} = \frac{56}{455}$

Simplify the fraction: To simplify $\frac{56}{455}$,

$\frac{56}{455} = \frac{8}{65}$

So, the probability that none of the balls drawn are red is:

$\frac{8}{65}$

Odds of an event

Odds show how likely or unlikely an event is to happen. They compare the chance of an event occurring to the chance of it not occurring.Odds are usually presented as a ratio.

For Example

• if the odds of winning a game are 2 to 1, it means there are two chances of winning for every one chance of losing.
• If the odds of rain tomorrow are 3 to 1, it means it's three times more likely to rain than not to rain.
• If the odds of flipping heads on a coin are 1 to 1, it means there's an equal chance of getting heads or tails.

Question 2. There are three events A, B and C of which only one must and one only one can happen. The odds are 8 to 3 against A, 5 to 2 against B. Find the odds agains C.

Solution:

Odd against A = 8 : 3
Odds in favour of A = 3 : 8,
P(A) = $\frac{3}{3+8} = \frac{3}{11}$

Odd against B = 5 : 2,
Odds in favour of B = 2 : 5,
P(B) = $\frac{2}{2+5} = \frac{2}{7}$

P(A) + P(B) + P(C) = 1
P(C) = 1 - P(A) - P(B)
P(C) = $1 - \frac{3}{11} - \frac{2}{7} = \frac{77 - 21 - 22}{77} = \frac{34}{77}$
Odds in favour of C = 34 : (77 - 34) = 34:43
Odds agains C = 43 : 34

Question 3. There are three events A, B and C of which only one must and one only one can happen. The odds are 8 to 3 against A, 5 to 2 against B. Find the odds agains C.

Solution:

Odd against A = 8 : 3
Odds in favour of A = 3 : 8,
P(A) = $\frac{3}{3+8} = \frac{3}{11}$

Odd against B = 5 : 2,
Odds in favour of B = 2 : 5,
P(B) = $\frac{2}{2+5} = \frac{2}{7}$

P(A) + P(B) + P(C) = 1
P(C) = 1 - P(A) - P(B)
P(C) = $1 - \frac{3}{11} - \frac{2}{7} = \frac{77 - 21 - 22}{77} = \frac{34}{77}$
Odds in favour of C = 34 : (77 - 34) = 34:43
Odds agains C = 43 : 34