*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 :Â **

The probability distribution of the random variable X is called aÂ **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, â€¦ n

p = the probability of success in a single trial

q = the probability of failure in a single trial

(i.e. q = 1 Ã¢Ë†â€™ p)

Keeping the above-mentioned idea in mind,

p is given to us as 3/10 or 0.3

=> q = 1 â€“ 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)^4Â = 15 * 0.09 * 0.2401 = 0.324135

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

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

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

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

As you can see from above,

total is =Â 0.117649Â +Â 0.302526Â +Â 0.324135Â +Â

0.18522Â + 0.059535Â +Â 0.010206Â +Â 0.000729Â = 1

(Add it up in calculator if you donâ€™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 â€“ P(5) + P(6)

=>Â 0.117649 + 0.302526 + 0.324135 + 0.18522 + 0.059535Â = 1 â€“ (0.010206 + 0.000729)

=> 0.989065 = 1 â€“ 0.010935

=>Â **Ans is ~ 0.99**