An MBM applies for a job in two firms X and Y. The probability is being selected in firm X is \[0.7\]and being rejected at Y is \[0.5\]. The probability of at least one of his applications being rejected is \[0.6\]. The probability that he will be selected in one of the firms, is
A. 0.6
B. 0.4
C. 0.8
D. None of these
Answer
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Hint: X And Y are two firms in which an MBM applied probability that may be selected in X is \[P(X) = 0.7\] probability that may not be selected in X is \[P(\mathop X\limits^ - ) = 1 - 0.7 = 0.3\]
Probability that may be selected in Y is \[P(Y) = 1 - P(Y)\]
\[\mathop Y\limits^ - \] denotes he may not be selected.
Complete step by step solution:
\[\begin{gathered}
P(X) = 0.7\,\,\,\,\,\,\,\,\,P(\mathop X\limits^ - ) = 0.3 \\
P(\mathop Y\limits^ - ) = 0.5\,\,\,\,\,\,\,\,\,\,\,P(Y) = 0.5 \\
and\,P(\mathop X\limits^ - \cup \mathop Y\limits^ - ) = 0.6 \\
\end{gathered} \]
Probability that person will be selected in one of the two firms
X or Y is
\[\begin{gathered}
P(X \cup Y) = 1 - P(\mathop X\limits^ - \cap \mathop Y\limits^ - ) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - (P(\mathop X\limits^ - ) + P(\mathop Y\limits^ - ) - P(\mathop X\limits^ - \cup \mathop Y\limits^ - ) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - (0.3 + 0.5 - 0.6) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - 0.2 = 0.8 \\
\end{gathered} \]
Thus, the correct answer is option C.
Additional Information. Probability is a branch of mathematics concerned with the analysis of random phenomena. The outcome of a random event cannot be determined before it occurs, but it may be any one of several possible outcomes. The actual outcome is considered to be determined by chance.
Experiment: Any phenomenon like rolling a dice, tossing a coin, drawing a card from a well-shuffled deck, etc.
Outcome: The Result of any event; like number appearing on a dice, side of a coin, drawn out card, etc.
Sample Space: The set of all possible outcomes.
Event: Any combination of possible outcomes or the subset of sample space; like getting an even number on rolled dice, getting a head/tail on a flipped coin, drawing out a king/queen/ace of any suit.
Probability Function: A function giving the probability for each outcome
Note: The solution can be done without using the formula i.e. By using the data Relationally.
Probability that may be selected in Y is \[P(Y) = 1 - P(Y)\]
\[\mathop Y\limits^ - \] denotes he may not be selected.
Complete step by step solution:
\[\begin{gathered}
P(X) = 0.7\,\,\,\,\,\,\,\,\,P(\mathop X\limits^ - ) = 0.3 \\
P(\mathop Y\limits^ - ) = 0.5\,\,\,\,\,\,\,\,\,\,\,P(Y) = 0.5 \\
and\,P(\mathop X\limits^ - \cup \mathop Y\limits^ - ) = 0.6 \\
\end{gathered} \]
Probability that person will be selected in one of the two firms
X or Y is
\[\begin{gathered}
P(X \cup Y) = 1 - P(\mathop X\limits^ - \cap \mathop Y\limits^ - ) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - (P(\mathop X\limits^ - ) + P(\mathop Y\limits^ - ) - P(\mathop X\limits^ - \cup \mathop Y\limits^ - ) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - (0.3 + 0.5 - 0.6) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - 0.2 = 0.8 \\
\end{gathered} \]
Thus, the correct answer is option C.
Additional Information. Probability is a branch of mathematics concerned with the analysis of random phenomena. The outcome of a random event cannot be determined before it occurs, but it may be any one of several possible outcomes. The actual outcome is considered to be determined by chance.
Experiment: Any phenomenon like rolling a dice, tossing a coin, drawing a card from a well-shuffled deck, etc.
Outcome: The Result of any event; like number appearing on a dice, side of a coin, drawn out card, etc.
Sample Space: The set of all possible outcomes.
Event: Any combination of possible outcomes or the subset of sample space; like getting an even number on rolled dice, getting a head/tail on a flipped coin, drawing out a king/queen/ace of any suit.
Probability Function: A function giving the probability for each outcome
Note: The solution can be done without using the formula i.e. By using the data Relationally.
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