The sample space for the experiment in question is shown in the table below. $\mathrm{S}=\left{\begin{array}{c}(3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \ 1...
Solution: Let $\mathrm{E}$ be the event that 'the sum of numbers on the dice is $4^{\prime}$ and $\mathrm{F}$ be the event that 'the two numbers appearing on throwing the two dice are different. So...
Solution: There are two sorts of questions in this section: True/False and Multiple Choice Questions (T/F or MCQ), and each of these types is further subdivided into Easy and Difficult categories,...
Solution Assume $B$ denotes boy and $G$ denote girl. Then, the sample space of the given experiment is $S={G G, G B, B G, B B}$ Let $\mathrm{E}$ be the event that 'both are girls'. $\Rightarrow...
and 
. Find (i) 
and 
and 
and 
Solution: The sample space for the given event is $S={1,2,3,4,5,6}$ Sample space of events,$E={1,3,5}, F={2,3}$ and $G={2,3,4,5} \ldots \ldots \ldots$ (i)The events probability of events are:...
(b) Find the conditional probability of obtaining the sum 8 , given that the red die resulted in a number less than 
Solution:Assuming: B denote black coloured die and R denote red coloured die. Therefore, the sample space for the specified experiment will be as follows:...
Solution: We assume that $\mathrm{M}$ denote mother, $\mathrm{F}$ denote father and $\mathrm{S}$ denote son. Then, the sample space for the given experiment will be: S= { MFS, SFM, FSM, MSF, SMF,...
Solution: We assume that $mathrm{M}$ denote mother, $mathrm{F}$ denote father and $mathrm{S}$ denote son. Then, the sample space for the given experiment will be: S= $\{$ MFS, SFM, FSM, MSF, SMF,...
Solution: We assume that $\mathrm{M}$ denote mother, $\mathrm{F}$ denote father and $\mathrm{S}$ denote son. Then, the sample space for the given experiment will be: S= ${$MFS, SFM, FSM, MSF, SMF,...
and 5 appears respectively on first two tosses.In the sample space, there are 216 outcomes, with each element of the sample space having three entries and taking the form $(x, y, z)$ where $1 \leq x, y, z \leq 6$. Considering the event, E: 4...
and 5 appears respectively on first two tosses.Solution: In the sample space, there are 216 outcomes, with each element of the sample space having three entries and taking the form $(x, y, z)$ where $\1 leq x, y, z leq 6$. Considering the event,...
and 5 appears respectively on first two tosses.Solution: In the sample space, there are 216 outcomes, with each element of the sample space having three entries and taking the form $(x, y, z)$ where $1 \leq x, y, z \leq 6$. Considering the...
: one coin shows head Solution: Determining the sample space of the given experiment is $\mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \Pi\}\}$ (i) Given that, E: tail appears on one coin And F: one coin shows head...
: head on third toss, 
: heads on first two tosses (ii) : at least two heads, 
: at most two heads (iii) 
: at most two tails, 
at least one tailSolution: The sample space of the given experiment will be: S={HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} (i) Given that, E: head-on third toss And F: heads on first two tossesSo writing the sample...
and 
, find
Solution: Given: $\mathrm{P}(\mathrm{A})=\frac{6}{11}, \mathrm{P}(\mathrm{B})=\frac{5}{11}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$ (i) We know that $\mathrm{P}\left(\mathrm{A}^{*}...
, if 
and 
.Solution: i) Given relation $2 \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})=\frac{5}{13}$ and $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{2}{5}$ $\Rightarrow \mathrm{P}(\mathrm{B})=\frac{5}{13},...
and 
, findSolution:Given $\mathrm{P}(\mathrm{A})=0.8, \mathrm{P}(\mathrm{B})=0.5$ and $\mathrm{P}(\mathrm{B} \mid \mathrm{A})=0.4$ (i) By definition of conditional probability, we know that,...
, if 
and 
Solution: Given: $P(B)=0.5$ and $P(A \cap B)=0.32$ We know this because conditional probability is defined as follows: $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap...
and are events such that 
and 
, find and Solution: Given: $P(E)=0.6, P(F)=0.3$ and $P(E \cap F)=0.2$ We know this because conditional probability is defined as follows: $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$ By substituting the values we...