Question

the probabilities of Abu , kuraku and musa passing their post jamb examination are 3/7 and 5/9 and 12/13 respectively . what is the probability that only one of them wiil pass. .......………………............... please help

Answer 1

The probability that only one of them will pass is 272/819

Explanation:

The probability that only one of them will pass is;

P(Abu pass, Kuraku and Musa failed) or P(Kuraku pass, Abu and Musa failed) or P(Musa passed and Abu and Kuraku failed)

Mathematically, the probability of an event occurring = (1- p(event not occurring))

What this means is that the probability of failing = 1 - probability of passing

So if P(A) = 3/7

P(A’) = 1-3/7 = 4/7

P(K) = 5/9, P(K’) = 1-5/9 = 4/9

P(M) = 12/13, P(M’) = 1-12/13 = 1/13

where A, K and M represents the names individually as their first letters and the ‘ represent the complement which is the probability of the events not happening

So the probability we want to calculate in notation form is;

P ={P(A) * P(K’) * P(M’)} + {P(A’) * P(K) * P(M’)} + {P(A’) * P(K’) * P(M)}

So substituting the values, we have;

P= {3/7 * 4/9 * 1/13} + {4/7 * 5/9 * 1/13} + {4/7 * 5/9 * 12/13} = 12/819 + 20/819 + 240/819 = 272/819