Questions & Answers

Question

Answers

$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P(A \cup B)$ | |

$(i)$ | $\dfrac{1}{3}$ | $\dfrac{1}{5}$ | $\dfrac{1}{{15}}$ | â€¦ |

$(ii)$ | $0.35$ | â€¦ | $0.25$ | $0.6$ |

$(iii)$ | $0.5$ | $0.35$ | â€¦ | $0.7$ |

Answer
Verified

Hint: In the above question we have to find some unknown values and some probability values are known to us. Use the basic formulae of \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\] to find out the value of these unknown quantities this will help to reach the right values of these quantities.

$(i)$ In the first part, we are given the values as

$P(A) = \dfrac{1}{3}$, $P(B) = \dfrac{1}{5}$,$P(A \cap B) = \dfrac{1}{{15}}$

We know that

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

Therefore, by using this formula and substituting the values of equation (1) in equation (2) we get

$P(A \cap B) = \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{{15}}$

$ = \dfrac{{5 + 3 - 1}}{{15}}$

$ = \dfrac{{8 - 1}}{{15}}$

$ = \dfrac{7}{{15}}$

$\therefore P(A \cap B) = \dfrac{7}{{15}}$

$(ii)$ In this part, we are given the values as

$P(A) = 0.35$,$P(A \cap B) = 0.25$,$P(A \cup B) = 0.6$ â€¦ (1)

We know that,

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\] â€¦ (2)

Therefore, by using this formula and substituting the values of equation (1) in equation (2) we get

\[0.6 = 0.35 + P(B) - 0.25\]

\[P(B) = 0.6 - 0.35 + 0.25\]

\[P(B) = 0.6 - 0.10\]

\[\therefore P(B) = 0.5\]

$(iii)$ In this part, we are given the values as

$P(A) = 0.5$, $P(B) = 0.35$,$P(A \cup B) = 0.7$

We know that,

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

Therefore, by using this formula and substituting the values of equation (1) in equation (2) we get

\[0.7 = 0.5 + 0.35 - P(A \cap B)\]

\[P(A \cap B) = 0.5 + 0.35 - 0.7\]

\[P(A \cap B) = 0.85 - 0.7\]

\[\therefore P(A \cap B) = 0.15\]

Therefore, after getting the required solutions, we can fill the table in the form:

Note: Whenever we face such types of problems the key point is to have a good grasp of the probability formula, some of them are mentioned above while performing the solution. Letâ€™s talk about the physical interpretation of $A \cup B$ this means that we have to find the probability of occurrence of event A and event B. This same concept can be applied to get the physical interpretation of $A \cap B$ as well.

$(i)$ In the first part, we are given the values as

$P(A) = \dfrac{1}{3}$, $P(B) = \dfrac{1}{5}$,$P(A \cap B) = \dfrac{1}{{15}}$

We know that

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

Therefore, by using this formula and substituting the values of equation (1) in equation (2) we get

$P(A \cap B) = \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{{15}}$

$ = \dfrac{{5 + 3 - 1}}{{15}}$

$ = \dfrac{{8 - 1}}{{15}}$

$ = \dfrac{7}{{15}}$

$\therefore P(A \cap B) = \dfrac{7}{{15}}$

$(ii)$ In this part, we are given the values as

$P(A) = 0.35$,$P(A \cap B) = 0.25$,$P(A \cup B) = 0.6$ â€¦ (1)

We know that,

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\] â€¦ (2)

Therefore, by using this formula and substituting the values of equation (1) in equation (2) we get

\[0.6 = 0.35 + P(B) - 0.25\]

\[P(B) = 0.6 - 0.35 + 0.25\]

\[P(B) = 0.6 - 0.10\]

\[\therefore P(B) = 0.5\]

$(iii)$ In this part, we are given the values as

$P(A) = 0.5$, $P(B) = 0.35$,$P(A \cup B) = 0.7$

We know that,

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

Therefore, by using this formula and substituting the values of equation (1) in equation (2) we get

\[0.7 = 0.5 + 0.35 - P(A \cap B)\]

\[P(A \cap B) = 0.5 + 0.35 - 0.7\]

\[P(A \cap B) = 0.85 - 0.7\]

\[\therefore P(A \cap B) = 0.15\]

Therefore, after getting the required solutions, we can fill the table in the form:

$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P(A \cup B)$ | |

$(i)$ | $\dfrac{1}{3}$ | $\dfrac{1}{5}$ | $\dfrac{1}{{15}}$ | $\dfrac{7}{{15}}$ |

$(ii)$ | $0.35$ | $0.5$ | $0.25$ | $0.6$ |

$(iii)$ | $0.5$ | $0.35$ | $0.15$ | $0.7$ |

Note: Whenever we face such types of problems the key point is to have a good grasp of the probability formula, some of them are mentioned above while performing the solution. Letâ€™s talk about the physical interpretation of $A \cup B$ this means that we have to find the probability of occurrence of event A and event B. This same concept can be applied to get the physical interpretation of $A \cap B$ as well.

×

Sorry!, This page is not available for now to bookmark.