91 questions · self-marked practice — reveal the answer and mark yourself.
| Assignment | $\omega_{1}$ | $\omega_{2}$ | $\omega_{3}$ | $\omega_{4}$ | $\omega_{5}$ | $\omega_{6}$ | $\omega_{7}$ |
| $\frac{1}{14}$ | $\frac{2}{14}$ | $\frac{3}{14}$ | $\frac{4}{14}$ | $\frac{5}{14}$ | $\frac{6}{14}$ | $\frac{15}{14}$ |
| Assignment | $\omega_{1}$ | $\omega_{2}$ | $\omega_{3}$ | $\omega_{4}$ | $\omega_{5}$ | $\omega_{6}$ | $\omega_{7}$ |
| -0.1 | 0.2 | 0.3 | 0.4 | -0.2 | 0.1 | 0.3 |
| P(A) | P(B) | P(A$\cap$B) | P(A$\cup$B) |
| 0.5 | 0.35 | ________ | 0.7 |
| P(A) | P(B) | P(A$\cap$B) | P(A$\cup$B) |
| 0.35 | _____ | 0.25 | 0.6 |
| P(A) | P(B) | P(A$\cap$B) | P(A$\cup$B) |
| $\frac 13$ | $\frac 15$ | $\frac 1{15}$ | _____ |
| Assignment | $\omega_{1}$ | $\omega_{2}$ | $\omega_{3}$ | $\omega_{4}$ | $\omega_{5}$ | $\omega_{6}$ | $\omega_{7}$ |
| 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 |
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\therefore \;0.8 = 0.5 + 0.4 - P(A \cap B)$
$\Rightarrow \;P(A \cap B) = 0.9 - 0.8 = 0.1$
$\therefore \;P(A \cap B) < P(A)\;{\text{and}}\;P(A \cap B) < P(B)$
Thus the given probabilities are consistently defined.
| Assignment | $\omega_{1}$ | $\omega_{2}$ | $\omega_{3}$ | $\omega_{4}$ | $\omega_{5}$ | $\omega_{6}$ | $\omega_{7}$ |
| $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ |
| Assignment | $\omega_{1}$ | $\omega_{2}$ | $\omega_{3}$ | $\omega_{4}$ | $\omega_{5}$ | $\omega_{6}$ | $\omega_{7}$ |
| 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
A or B =$A \cup B$ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, l), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = S
A' = S - A
= (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} = B
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let event A: getting three heads = {HHH},
event B: getting exactly two heads = {HHT, HTH, THH} and
event C: getting three tails = {TTT}
Now $\style{font-size:28px}{A\cap B=\phi}$, $\style{font-size:28px}{B\cap C=\phi}$ and $\style{font-size:28px}{A\cap C=\phi}$
Thus A, B, C are mutually exclusive events
Also ${A\cup B\cup C=\{HHH,\;HHT,\;HTH,\;THH,\;TTT\;\}⧧S}$
Thus A, B, C are mutually exclusive but not exhaustive events.
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Now, let A be the event: getting at least two heads = {HHH, HHT, HTH, THH}
B be the event: getting exactly one head = {HTT, THT, TTH}
and C be th event: getting no head = {TTT}
Mutually exclusive events are those in which no element is common
Since ${A\cap B=\phi}$. $\therefore$ A and B are mutually exclusive events,
$\style{font-size:28px}{B\cap C=\phi}$. $\therefore$ B and C are mutually exclusive events,
$\style{font-size:28px}{A\cap C=\phi}$. $\therefore$ A and C are mutually exclusive events.
Thus A, B, C are mutually exclusive events.
Also, as ${A\cup B\cup C=\{HHH,\;HHT,\;HTH,\;THH,\;HTT,\;THT,\;TTH,\;TTT\}=S}$
So, A, B, C are exhaustive events.
A, B, C are three events which are mutually exclusive and exhaustive.
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Event A: three heads show = {HHH}
Event B: two heads and one tail show = {HHT, HTH, THH}
Event C: three tails show = {TTT}
Event D: a head shows on the first coin - {HHH, HHT, HTH, HTT}
A compound event is the occurrence of two or more outcomes together.
We see that, B = {HHT, HTH, THH}, and D = {HHH, HHT, HTH, HTT}.
Since cardinal numbers of events B and D are respectively 3 and 4,
So, B and D are compound events.
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Event A: three heads show = {HHH}
Event B: two heads and one tail show = {HHT, HTH, THH}
Event C: three tails show = {TTT}
Event D: a head shows on the first coin - {HHH, HHT, HTH, HTT}
Simple events have only one outcome
A ={HHH}, Here cardinal number = 1
$\therefore$ A is a simple event.
Since, C = {TTT}, Here also cardinal number = 1
$\therefore$ C is a simple event.
| Outcomes | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
| 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
| Outcomes | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
| $\frac 1{12}$ | $\frac 1{12}$ | $\frac 16$ | $\frac 16$ | $\frac 16$ | $\frac 32$ |
| Outcomes | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
| $\frac 18$ | $\frac 23$ | $\frac 13$ | $\frac 13$ | $-\frac 14$ | $-\frac 13$ |
| Outcomes | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
| 1 | 0 | 0 | 0 | 0 | 0 |
| Outcomes | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
| $\frac 16$ | $\frac 16$ | $\frac 16$ | $\frac 16$ | $\frac 16$ | $\frac 16$ |
Therefore, total number of outcomes, n (S) = 24.
Sample space for the experiment is given by
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let the event ‘she visits A before B’ be denoted by E
Therefore, E = {ABCD, CABD, DABC, ABDC, CADB, DACB
ACBD, ACDB, ADBC, CDAB, DCAB, ADCB}
n(E) =12
Thus P(E) = $\frac {n(E)}{n(S)} =\frac {12}{24} = \frac 12$