Let A be the event in which the selected student has opted for NCC and B be the event in which the selected student has opted for NSS.
Total number of students = 60
Number of students who have opted for NCC = 30
\(∴P(A)=\frac{30}{60}=\frac{1}{2}\)
Number of students who have opted for NSS = 32
\(∴P(B)=\frac{32}{60}=\frac{8}{15}\)
Number of students who have opted for both NCC and NSS = 24
\(∴P(A\) and \(B)=\frac{24}{60}=\frac{2}{5}\)
(i) We know that\(P(A\)or \(B) = P(A) + P(B) - P(A\) and \(B)\)
\(∴P(A\)or\(B)=\frac{1}{2}+\frac{8}{15}-\frac{2}{5}=15+16-\frac{12}{30}=\frac{19}{30}\)
Thus, the probability that the selected student has opted for NCC or NSS is \(\frac{19}{30}.\)
(ii)P(not A and not B)
=P(A' and B')
=P(A' ∩ B')
=P(AUB)' [(A'∩B')=(AUB)' (by de morgan's law]
\(=1-P(AUB)\)
\(=1-P(A\) or \(B)\)
\(=1-\frac{19}{30}\)
\(=\frac{11}{30}\)
Thus, the probability that the selected students has neither opted for NCC nor NSS is \(\frac{11}{30}.\)
(iii) The given information can be represented by a Venn diagram as
It is clear that Number of students who have opted for NSS but not NCC
\(= n(B - A) = n(B) - n(A ∩ B) = 32 - 24 = 8\)
Thus, the probability that the selected student has opted for NSS but not for NCC =\(\frac{8}{60}=\frac{2}{15}.\)
Let AAA be the event where a selected student opted for NCC, and B be the event where a selected student opted for NSS.
Total number of students: 60
The probability that a student opted for NCC, \(P(A) = \frac{30}{60} = \frac{1}{2}\)
The probability that a student opted for NSS, \(P(B) = \frac{32}{60} = \frac{8}{15}\)
The probability that a student opted for both NSS and NCC, \(P(A \cap B) = \frac{24}{60} = \frac{2}{5}\)
(i) The probability that a student opted for either NSS or NCC,
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(P(A \cup B) = \frac{1}{2} + \frac{8}{15} - \frac{2}{5}\)
\(P(A \cup B) = \frac{15}{30} + \frac{16}{30} - \frac{12}{30}\)
\(P(A \cup B) = \frac{19}{30}\)
(ii) The probability that a student opted for neither NSS nor NCC,
\(P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B)\)
\(P(A' \cap B') = 1 - \frac{19}{30}\)
\(P(A' \cap B') = \frac{11}{30}\)
(iii) Number of students who opted for NSS but not NCC,
\(n(B - A) = n(B) - n(A \cap B)\)
\( n(B−A)=32−24\)
\( n(B−A)=8\)
The probability that a selected student opted for NSS but not NCC,\(P(B - A) = \frac{8}{60} = \frac{2}{15}\)
Therefore:
Three coins are tossed once. Find the probability of getting
(i) 3 heads (ii) 2 heads (iii) at least 2 heads
(iv) at most 2 heads (v) no head (vi) 3 tails
(vii) exactly two tails (viii) no tail (ix) at most two tails
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
The axiomatic probability perspective is a unifying perspective in which the coherent conditions used in theoretical and experimental probability exhibit subjective probability. Kolmogorov's set of rules or axioms is put to all types of probability. They are known as Kolmogorov's 3 axioms by mathematicians. You can use axiomatic probability to calculate the likelihood of an event that is occurring or not occurring.
The 3 axioms are applicable to all other probability perspectives. This viewpoint is defined as the probability of any function from numbers to events that are satisfied by the three axioms listed below: