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Three couples and two single individuals have been invited to an investment seminar. Let X to be the number of people who arrive late for the seminar.

The probability of any individual or couples arrives late at 0.40. It can be represented by p. Similarly, probability of not arriving late is represented by q. So, the probability of arrival on time of couple or an individual is computed as:

\(\begin{aligned}q &= 1 - p\\ &= 1 - 0.40\\ &= 0.60\end{aligned}\)

It is given that the random variableX to be the number of people who arrive late for the seminar.

So, the probability that no one is late i.e.\(P\left( {X = 0} \right)\)is computed as:

\(\begin{aligned}P\left( {No\,one\,is\,late} \right) &= P\left( {X = 0} \right)\\ &= {\left( {0.6} \right)^5}\\ &= 0.0778\end{aligned}\)

Similarly, the probability that one individual is late i.e.\(P\left( {X = 1} \right)\)is computed as:

\(\begin{aligned}P\left( {one\,individual\,late} \right) &= P\left( {X = 1} \right)\\ &= {}^2{C_1}{\left( {0.40} \right)^1}{\left( {0.60} \right)^{5 - 1}}\\ &= \frac{{2!}}{{1!\left( {2 - 1} \right)!}} \times \left( {0.40} \right) \times {\left( {0.60} \right)^4}\\ &= 0.1037\end{aligned}\)

Similarly, the probability that two individuals or one couple are late i.e.\(P\left( {X = 2} \right)\)is computed as:

\(\begin{aligned}P\left( {two\,individual\,late} \right) + P\left( {1\,couple} \right) &= P\left( {X = 2} \right)\\ &= {\left( {0.40} \right)^2}{\left( {0.60} \right)^3} + {}^3{C_1}{\left( {0.40} \right)^1}{\left( {0.60} \right)^{5 - 1}}\\ &= {\left( {0.40} \right)^2}{\left( {0.60} \right)^3} + \frac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \left( {0.40} \right) \times {\left( {0.60} \right)^4}\\ &= 0.1901\end{aligned}\)

Similarly, the probability that one individual and one couple are late i.e.\(P\left( {X = 3} \right)\)is computed as:

\(\begin{aligned}P\left( {1\,\,individual\,late} \right)\,\,and\,P\left( {1\,\,couple} \right) &= P\left( {X = 3} \right)\\ &= \left( {{}^2{C_1}{{\left( {0.40} \right)}^1}} \right)\left( {{}^3{C_1}{{\left( {0.40} \right)}^1}{{\left( {0.60} \right)}^3}} \right)\\ &= 0.2074\end{aligned}\)

Similarly, the probability that two individual and one couple are late or 2 couples are late i.e.\(P\left( {X = 4} \right)\)is computed as:

\(\begin{aligned}P\left( {2\,\,individuals\,\,and\,\,1\,\,couple\,late} \right)\,\, + \,P\left( {2\,\,couples} \right) &= P\left( {X = 4} \right)\\ &= \left( \begin{array}{l}{\left( {0.4} \right)^2}\left( {{}^3{C_1}{{\left( {0.40} \right)}^1}{{\left( {0.60} \right)}^2}} \right)\\ + \left( {{}^3{C_2}{{\left( {0.40} \right)}^2}{{\left( {0.60} \right)}^3}} \right)\end{array} \right)\\ &= 0.1728\end{aligned}\)

Similarly, the probability that one individual and two couples are late i.e.\(P\left( {X = 5} \right)\)is computed as:

\(\begin{aligned}P\left( {1\,\,individual\,and\,2\,\,couples\,late} \right)\,\,& = P\left( {X = 5} \right)\\ &= \left( {{}^2{C_1}{{\left( {0.40} \right)}^1}} \right)\left( {{}^3{C_2}{{\left( {0.40} \right)}^2}{{\left( {0.60} \right)}^4}} \right)\\ &= 0.1382\end{aligned}\)

Similarly, the probability that two individuals and two couples are late or 3 couples are late i.e.\(P\left( {X = 6} \right)\)is computed as:

\(\begin{aligned}P\left( {2\,individuals\,and\,2\,couples\,late} \right)\, + P\left( {3\,couples} \right) &= P\left( {X = 6} \right)\\\left( \begin{array}{l}{\left( {0.4} \right)^2}\left( {{}^3{C_2}{{\left( {0.40} \right)}^2}\left( {0.60} \right)} \right)\\ + {\left( {0.40} \right)^3}{\left( {0.60} \right)^2}\end{array} \right)\\ &= 0.0691\end{aligned}\)

Similarly, the probability that one individual and three couples are late i.e.\(P\left( {X = 7} \right)\)is computed as:

\(\begin{aligned}P\left( {1\,\,{\rm{individual}}\,\,{\rm{and}}\,\,3\,{\rm{couples}}\,{\rm{late}}} \right)\, &= P\left( {X = 7} \right)\\ &= \left( {{}^2{C_1}{{\left( {0.40} \right)}^1}{{\left( {0.40} \right)}^3}\left( {0.60} \right)} \right)\\ &= 0.0307\end{aligned}\)

Finally, the probability that two individuals and three couples are late i.e.\(P\left( {X = 8} \right)\)is computed as:

\(\begin{aligned} P\left( {all\,late} \right)\, &= P\left( {X = 8} \right)\\ &= {\left( {0.40} \right)^5}\\ &= 0.0102\end{aligned}\)

The probability mass function of X is given as:

From the problem, the cdf\(F\left( {X = x} \right)\) can be defined as,

\(F\left( {X = x} \right) = P\left( {X \le x} \right)\)

First, compute\(F\left( 0 \right)\) using the above formula:

\(\begin{aligned} F\left( {X = 0} \right) &= P\left( {X \le 0} \right)\\ &= P\left( {X = 0} \right)\\ &= 0.0776\end{aligned}\)

Then, compute\(F\left( 1 \right)\) using the above formula:

\(\begin{aligned} F\left( {X = 1} \right) &= P\left( {X \le 1} \right)\\ &= \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right)} \right)\\ &= \left( {0.0776 + 0.1037} \right)\\ &= 0.1813\end{aligned}\)

Then, compute\(F\left( 2 \right)\) using the above formula:

\(\begin{aligned} F\left( {X = 2} \right) &= P\left( {X \le 2} \right)\\ &= \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right)} \right)\\&= \left( {0.0776 + 0.1037 + 0.1901} \right)\\ &= 0.3714\end{aligned}\)

Similarly, compute\(F\left( 3 \right)\) using the above formula:

\(\begin{aligned} F\left( {X = 3} \right) &= P\left( {X \le 3} \right)\\ &= \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) + P\left( {X = 3} \right)} \right)\\ &= \left( {0.0776 + 0.1037 + 0.1901 + 0.2074} \right)\\ &= 0.5788\end{aligned}\)

Similarly, compute\(F\left( 4 \right)\) using the above formula:

\(\begin{aligned}F\left( {X = 4} \right) &= P\left( {X \le 4} \right)\\ &= \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) + P\left( {X = 3} \right) + P\left( {X = 4} \right)} \right)\\ &= \left( {0.0776 + 0.1037 + 0.1901 + 0.2074 + 0.1728} \right)\\ &= 0.7516\end{aligned}\)

Similarly, compute\(F\left( 5 \right)\) using the above formula:

\(\begin{aligned} F\left( {X = 5} \right) &= P\left( {X \le 5} \right)\\ &= \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) + P\left( {X = 3} \right) + P\left( {X = 4} \right) + P\left( {X = 5} \right)} \right)\\ &= \left( {0.0776 + 0.1037 + 0.1901 + 0.2074 + 0.1728 + 0.1382} \right)\\ &= 0.8898\end{aligned}\)

Similarly, compute\(F\left( 6 \right)\) using the above formula:

\(\begin{aligned} F\left( {X = 6} \right) &= P\left( {X \le 6} \right)\\ &= \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) + P\left( {X = 3} \right) + P\left( {X = 4} \right) + P\left( {X = 5} \right) + P\left( {X = 6} \right)} \right)\\ &= \left( {0.0776 + 0.1037 + 0.1901 + 0.2074 + 0.1728 + 0.1382 + 0.0693} \right)\\ &= 0.9591\end{aligned}\)

Similarly, compute\(F\left( 7 \right)\) using the above formula:

\(\begin{aligned} F\left( {X = 7} \right) &= P\left( {X \le 7} \right)\\ &= \left( \begin{array}{l}P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) + P\left( {X = 3} \right)\\ + P\left( {X = 4} \right) + P\left( {X = 5} \right) + P\left( {X = 6} \right) + P\left( {X = 6} \right)\end{array} \right)\\ &= \left( {0.0776 + 0.1037 + 0.1901 + 0.2074 + 0.1728 + 0.1382 + 0.0693 + 0.0307} \right)\\ &= 0.9898\end{aligned}\)

Similarly, compute\(F\left( 8 \right)\) using the above formula:

\(\begin{aligned} F\left( {X = 8} \right) &= P\left( {X \le 8} \right)\\ &= \left( \begin{array}{l}P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) + P\left( {X = 3} \right) + P\left( {X = 4} \right)\\ + P\left( {X = 5} \right) + P\left( {X = 6} \right) + P\left( {X = 7} \right) + P\left( {X = 8} \right)\end{array} \right)\\ &= \left( \begin{array}{l}0.0776 + 0.1037 + 0.1901 + 0.2074 + 0.1728\\ + 0.1382 + 0.0693 + 0.0307 + 0.0102\end{array} \right)\\ &= 1\end{aligned}\)

The cumulative distribution functionis given as:

The probability can be computed as:

\(\begin{aligned}P\left( {2 \le X \le 6} \right) &= F\left( 6 \right) - F\left( 1 \right)\\ &= 09591 - 0.1813\\ &= 0.7778\end{aligned}\)

Thus, the required probability is 0.7778.