*Wednesday, January 22nd, 2014*

**Question : Six persons try to swim across a wide river.Its known that on an average only three persons out of 10 are successful in crossing the river.Whats the probability that at most four of the 6 persons will cross the river safely?**

**Answer : **

The probability distribution of the random variable X is called a **binomial distribution**, and is given by the formula:

* P(X)= nCr * p^r * q^(n-r)
where
n = the number of trials
r = 0, 1, 2, … n
p = the probability of success in a single trial
q = the probability of failure in a single trial
(i.e. q = 1 âˆ’ p)*

Keeping the above mentioned idea in mind,

p is given to us as 3/10 or 0.3

=> q = 1 – 0.3 = 0.7

Also, n = 6 {There are 6 people trying to cross the river}

P(0) = 6C0 * (0.3)^0 * (0.7)^6 = 1 * 1 * 0.117649 = 0.117649

P(1) = 6C1 * (0.3)^1 * (0.7)^5 = 6 * 0.3 * 0.16807 = 0.302526

P(2) = 6C2 * (0.3)^2 * (0.7)^4 = 15 * 0.09 * 0.2401 = 0.324135

P(3) = 6C3 * (0.3)^3 * (0.7)^3 = 20 * 0.027 * 0.343 = 0.18522

P(4) = 6C4 * (0.3)^4 * (0.7)^2 = 15 * 0.0081 * 0.49 = 0.059535

P(5) = 6C5 * (0.3)^5 * (0.7)^1 = 6 * 0.00243 * 0.7 = 0.010206

P(6) = 6C6 * (0.3)^6 * (0.7)^0 = 1 * 0.000729 * 1 = 0.000729

As you can see from above, total is = 0.117649 + 0.302526 + 0.324135 + 0.18522 +0.059535 + 0.010206 + 0.000729 = 1 {Add it up in calculator if you don’t believe me }

The question is asking us for at most 4 crosses, so that would be

P(0) + P(1) + P(2) + P(3) + P(4) = 1 – P(5) + P(6)

=> 0.117649 + 0.302526 + 0.324135 + 0.18522 + 0.059535 = 1 – (0.010206 + 0.000729)

=> 0.989065 = 1 – 0.010935

=> **Ans is ~ 0.99**