# The probability of winning one bond of a 3-percent loan is 0.25. Find the probability that out

**The probability of winning one bond of a 3-percent loan is 0.25. Find the probability that out of 8 purchased bonds, the following will be winning: a) three; b) two; c) at least two**

We use Bernoulli’s formula (determining the probability of occurrence of event A exactly m times (no matter in what order) with probability p in one outcome from a series of n experiments) to calculate all options.

p m, n = Cmn * p m * (1 – p) n-m, where

Cmn = n! / (m! * (n – m)!)

a) event A consists in the fact that out of 8 bonds, 3 are winning and 5 are not winning, that is, n = 8, m = 3).

C38 = 8! / (3! * (8 – 3)!) = 40320 / (6 * 120) = 40320/720 = 56

p (A) = C38 * p3 * (1 – p) 8-3 = 56 * 0.253 * 0.755 = 56 * 0.016 * 0.237 = 0.212

b) event B consists in the fact that out of 8 bonds, 2 are winning and 6 are not winning, that is, n = 8, m = 2).

C28 = 8! / (2! * (8 – 2)!) = 40320 / (2 * 720) = 40320/1440 = 28

p (B) = C28 * p2 * (1 – p) 8-2 = 28 * 0.252 * 0.756 = 28 * 0.063 * 0.178 = 0.314

c) event B is that there should be at least two winning bonds, that is, any outcome will suit, except for the outcome when none of the bonds wins or only one. To do this, you need to calculate the probabilities of the remaining events and then add them up. But it is better to use the formula for the probability of the reverse event:

p (B) = 1 – (p (T) + p (D)), where

р (Г) is the probability of winning one bond,

p (D) is the probability of losing all bonds.

Let’s calculate p (G):

C18 = 8! / (1! * (8 – 1)!) = 40320 / (1 * 5040) = 40320/5040 = 8

p (G) = C18 * p * (1 – p) 8-1 = 8 * 0.25 * 0.757 = 8 * 0.25 * 0.133 = 0.266

Let’s calculate p (D):

C08 = 8! / (0! * (8 – 0)!) = 40320 / (1 * 40320) = 40320/40320 = 1

p (D) = C08 * p0 * (1 – p) 8-0 = 1 * 0.250 * 0.758 = 1 * 1 * 0.100 = 0.100

Eventually:

P (B) = 1 – (0.266 + 0.100) = 1 – 0.366 = 0.634

Answer: a) 0.212; b) 0.314; c) 0.634.