Exercise 8.4 - Chapter 8 Statistics 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $8.4$
Question 1.

If $\mathrm{P}(\mathrm{A})=\frac{2}{3} \mathrm{P}(\mathrm{B})=\frac{2}{\frac{2}{5}}(\mathrm{~A} \cup \mathrm{B})=13$ then find $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$.
Solution:
$
\begin{aligned}
& \mathrm{P}(\mathrm{A})=\frac{2}{3}, \mathrm{P}(\mathrm{B})=\frac{2}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{1}{3} \\
& \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\
& =\frac{2}{3}+\frac{2}{5}-\frac{1}{3} \\
& =\frac{10+6-5}{15}=\frac{11}{15}
\end{aligned}
$
Question 2.
$\mathrm{A}$ and $\mathrm{B}$ are two events such that, $\mathrm{P}(\mathrm{A})=0.42, \mathrm{P}(\mathrm{B})=0.48$, and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.16$. Find
(i) $\mathrm{P}(\operatorname{not} \mathrm{A})$
(it) $\mathrm{P}($ not $B)$
(iii) $P(A$ or $B)$
Answer:
(i) $\mathrm{P}($ not $\mathrm{A})=1-\mathrm{P}(\mathrm{A})=1-0.42=0.58$
(ii) $\mathrm{P}($ not $\mathrm{B})=1-\mathrm{P}(\mathrm{B})=1-0.48=0.52$
(iii) $\mathrm{P}(\mathrm{A}$ or $\mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.42+0.48-0.16=0.90-0.16=0.74$
Question 3.
If $\mathrm{A}$ and $\mathrm{B}$ are two mutually exclusive events of a random experiment and $\mathrm{P}($ not $\mathrm{A})=0.45$, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.65$, then find $\mathrm{P}(\mathrm{B})$.
Solution:
A and B are two mutually exclusive events of a random experiment.
$\mathrm{P}($ not $\mathrm{A})=0.45$
$
\mathrm{P}(\mathrm{A})=1-\mathrm{P}(\text { not } \mathrm{A})
$

$
\begin{aligned}
& \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.65=1-0.45=0.55 \\
& \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=0.65 \\
& 0.55+\mathrm{P}(\mathrm{B})=0.65 \\
& \mathrm{P}(\mathrm{B})=0.65-0.55 \\
& =0.10
\end{aligned}
$
Question 4.
The probability that atleast one of $\mathrm{A}$ and $\mathrm{B}$ occur is $0.6$. If $\mathrm{A}$ and $\mathrm{B}$ occur simultaneously with probability $0.2$, then find $\mathrm{P}(\bar{A})+\mathrm{P}(\bar{B})$.
Answer:
Here $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.6, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$0.6=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-0.2$
$\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=0.8$
$\mathrm{P}(\bar{A})+\mathrm{P}(\bar{B})=1-\mathrm{P}(\mathrm{A})+1-\mathrm{P}(\mathrm{B})$
$=2-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})]$
$=2-0.8$
$=1.2$
Question 5.
The probability of happening of an event $\mathrm{A}$ is $0.5$ and that of $\mathrm{B}$ is $0.3$. If $\mathrm{A}$ and $\mathrm{B}$ are mutually exclusive events, then find the probability that neither A nor B happen.
Solution:
$\mathrm{P}(\mathrm{A})=0.5$ Since $\mathrm{A}$ and $\mathrm{B}$ are mutually inclusive events
$\mathrm{P}(\mathrm{B})=0.3$ events.
$\mathrm{P}(\overline{\mathbf{A}}) \cup \mathrm{P}(\overline{\mathbf{B}})=1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})]$
$=1-[0.5+0.3]=0.2$
Question 6.
Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.
Solution:
Two dice rolled once.

$
\begin{aligned}
& n(\mathrm{~S})=36 \\
&
\end{aligned}
$
Happening of an even number in the first die is $\mathrm{A}$.
$
\begin{aligned}
A= & \left\{\begin{array}{l}
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
\end{array}\right\} \\
\Rightarrow \quad & n(\mathrm{~A})=18 \\
& \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{18}{36}
\end{aligned}
$

Happening of a total of face sum is 8 is $B$.
$
\begin{aligned}
\mathrm{B} & =\{(2,6),(3,5),(4,4),(5,3),(6,2)\} \\
n(\mathrm{~B}) & =5 \\
\mathrm{P}(\mathrm{B}) & =\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{5}{36} \\
(\mathrm{~A} \cap \mathrm{B}) & =\{(2,6),(4,4),(6,2)\} \\
n(\mathrm{~A} \cap \mathrm{B}) & =3 \\
\mathrm{P}(\mathrm{A} \cap \mathrm{B}) & =\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{3}{36} \\
\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B}) & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
, & \frac{18}{36}+\frac{5}{36}-\frac{3}{36}=\frac{18+5-3}{36} \\
& =\frac{20}{36}=\frac{5}{9}
\end{aligned}
$
Question 7.
From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of it being either a red king or a black queen.
Solution:
$
\mathrm{n}(\mathrm{S})=52
$
No. of Red cards $=26$,
Red king cards $=2$
No. of Black cards $=26$,
Black queen cards $=2$
No. of red king cards $=n(K)=2$

$
\therefore \mathrm{P}(\mathrm{K})=\frac{n(\mathrm{~K})}{n(\mathrm{~S})}=\frac{2}{52}
$
No. of black queen cards $n(\mathrm{Q})=2$
$
\begin{aligned}
\mathrm{P}(\mathrm{Q}) & =\frac{n(\mathrm{Q})}{n(\mathrm{~S})}=\frac{2}{52} \\
\therefore \quad n(\mathrm{~K} \cap \mathrm{Q}) & =0 \\
\therefore \quad \mathrm{P}(\mathrm{K} \cup \mathrm{Q}) & =\mathrm{P}(\mathrm{K})+\mathrm{P}(\mathrm{Q})-\mathrm{P}(\mathrm{K} \cap \mathrm{Q}) \\
& =\frac{2}{52}+\frac{2}{52}-0 \\
& =\frac{4}{52}=\frac{1}{13}
\end{aligned}
$
$\therefore$ The probability of being either a red king or a black queen $=\frac{1}{13}$.
Question 8.
A box contains cards numbered $3,5,7,9, \ldots 35,37$. A card is drawn at random from the box. Find the probability that the drawn card has either multiples of 7 or a prime number.
Solution:
$
\begin{aligned}
& \mathrm{S}=\{3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37\} \\
& \mathrm{n}(\mathrm{S})=18 \\
& \text { Multiplies of seven cards }(\mathrm{A})=\{7,21,35\} \\
& =\mathrm{n}(\mathrm{A})=3 \\
& \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{3}{18}
\end{aligned}
$
Let the prime number cards B
$
\mathrm{B}=\{3,5,7,11,13,17,19,23,29,31,37\}
$

$
\begin{aligned}
& \mathrm{n}(\mathrm{B})=11 \\
& \mathrm{P}(\mathrm{B})=\frac{11}{18} \\
& \therefore(\mathrm{A} \cap \mathrm{B})=\{7\} \\
& n(\mathrm{~A} \cap \mathrm{B})=1 \\
& \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{18} \\
& \therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
&=\frac{3}{18}+\frac{11}{18}-\frac{1}{18}=\frac{13}{18}
\end{aligned}
$
$\therefore$ Probability of the drawn card is either
$
\text { multiples of seven or a prime number }=\frac{13}{18}
$
Question 9.
Three unbiased coins are tossed once. Find the probability of getting at most 2 tails or at least 2 heads.
Solution:
When we toss three coins, the sample space $\mathrm{S}=\{\mathrm{HHH}$, TTT, HTT, THH, HHT, TTH, HTH, THT $\}$ $\mathrm{n}(\mathrm{S})=8$
Event of getting at most 2 tails be $\mathrm{A}$.

$\therefore \mathrm{A}=\{\mathrm{HHH}, \mathrm{HTT}, \mathrm{THH}, \mathrm{HHT}, \mathrm{TTH}, \mathrm{HTH}, \mathrm{THT}\}$
$
\begin{aligned}
& n(\mathrm{~A})=7 \\
& \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{7}{8}
\end{aligned}
$
Event of getting atleast 2 heads be B.
$
\begin{aligned}
& \therefore \mathrm{B}=\{\mathrm{HHH}, \mathrm{THH}, \mathrm{HHT}, \mathrm{HTH}\} \\
& n(\mathrm{~B})=4 \\
& P(B)=\frac{4}{8} \\
& \mathrm{~A} \cap \mathrm{B}=\{\mathrm{HHH}, \mathrm{THH}, \mathrm{HHT}, \mathrm{HTH}\} \\
& n(\mathrm{~A} \cap \mathrm{B})=4, \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& =\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{4}{8} \\
& \therefore \quad \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& =\frac{7}{8}+\frac{4}{8}-\frac{4}{8}=\frac{7}{8} \\
&
\end{aligned}
$
Question 10.
The probability that a person will get an electrification contract is the probability that he will not get plumbing contract is $\frac{3}{5}$. The probability of getting at least one contract is $\frac{5}{8}$. What is the probability that he will get both?
Solution:

Let $\mathrm{P}(\mathrm{A})=\frac{3}{5}$
$\mathrm{P}(\mathrm{B})=1-\frac{5}{8}=\frac{3}{8}$
$
\begin{aligned}
\mathrm{P}(\mathrm{A} \cup \mathrm{B}) & =\frac{5}{7} \\
\mathrm{P}(\mathrm{A} \cap \mathrm{B}) & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\
& =\frac{3}{5}+\frac{3}{8}-\frac{5}{7} \\
& =\frac{168+105-200}{280}=\frac{73}{280}
\end{aligned}
$
Probability of getting both offer $=\frac{73}{280}$
Question 11.
In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that $30 \%$ of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years ?

Solution:
$
n(\mathrm{~S})=8000
$
Over 50 years be $\mathrm{A} ; n(\mathrm{~A})=1300$
Females be $\mathrm{B} ; \quad n(\mathrm{~B})=3000$ $30 \%=\frac{30}{100}$ of 3000 are over 50 years.
$
\begin{aligned}
\text { i.e. } \frac{30}{100} \times 3000 & =900 \\
n(\mathrm{~A} \cap \mathrm{B}) & =900 \\
\therefore \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})} & =\frac{1300}{8000}, \mathrm{P}(\mathrm{B})=\frac{3000}{8000}, \\
\mathrm{P}(\mathrm{A} \cap \mathrm{B}) & =\frac{900}{8000} \\
\therefore \quad \mathrm{P}(\mathrm{A} \cup \mathrm{B}) & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& =\frac{1300}{8000}+\frac{3000}{8000}-\frac{900}{8000} \\
& =\frac{3400}{8000}=\frac{17}{40}=0.425
\end{aligned}
$

Question 12.
A coin is tossed thrice. Find the probability of getting exactly two heads or at least one tail or two consecutive heads.
Solution:
Three coins tossed simultaneously.
$\mathrm{S}=\{$ HHH, TTT, HHT, TTH, HTH, THT, HTT, THH $\}$
$\mathrm{n}(\mathrm{S})=8$
Happening of getting exactly two heads be A.
$\mathrm{A}=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$
$\mathrm{n}(\mathrm{A})=3$
$
\mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{3}{8}
$
Event of getting at least one tail be $B$.
$\therefore \mathrm{B}=\{\mathrm{TTT}, \mathrm{HHT}, \mathrm{TTH}, \mathrm{HTH}, \mathrm{THT}, \mathrm{HTT}, \mathrm{THH}\}$

$
\begin{aligned}
& n(\mathrm{~B})=7 \\
& \mathrm{P}(\mathrm{B})=\frac{7}{8}
\end{aligned}
$
Event of getting consecutively two heads be $\mathrm{C}$.
$
\begin{aligned}
& \mathrm{C}=\{\mathrm{HHT}, \mathrm{THH}, \mathrm{HHH}\} \\
& n(\mathrm{C})=3 \\
& \mathrm{P}(\mathrm{C})=\frac{3}{8} \\
& \mathrm{~A} \cap \mathrm{C}=\{\mathrm{HHT}, \mathrm{THH}\} \\
& n(\mathrm{~A} \cap \mathrm{C})= 2, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{2}{8} \\
& \mathrm{~A} \cap \mathrm{B}=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\} \\
& n(\mathrm{~A} \cap \mathrm{B})= 3, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{3}{8} \\
& \mathrm{~B} \cap \mathrm{C}=\{\mathrm{HHT}, \mathrm{THH}\} \\
& n(\mathrm{~B} \cap \mathrm{C})= 2, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{2}{8} \\
& \therefore(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=\{\mathrm{HHT}, \mathrm{THH}\} \\
& n(\mathrm{~A} \cap \mathrm{B} \cap \mathrm{C})= 2 \\
& \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})= \frac{2}{8} \\
& \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})= \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})- \\
& \mathrm{P}(\mathrm{B} \cap \mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{C})+\mathrm{P}(\mathrm{A} \cap \\
&\mathrm{B} \cap \mathrm{C})
\end{aligned}
$

$=\frac{3}{8}+\frac{7}{8}+\frac{3}{8}-\frac{3}{8}-\frac{2}{8}-\frac{2}{8}+\frac{2}{8}=\frac{8}{8}=1$

Question 13.
If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are any three events such that probability of $\mathrm{B}$ is twice as that of probability of $\mathrm{A}$ and probability of $\mathrm{C}$ is thrice as that of probability of $\mathrm{A}$ and if $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{4}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=$ $\frac{1}{8}, \mathbf{P}(\mathbf{A} \cup \mathbf{B} \cup \mathbf{C})=\frac{9}{10}, \mathbf{P}(\mathbf{A} \cap \mathbf{B} \cap \mathbf{C})=\frac{1}{15}$, then find $\mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B})$ and $\mathrm{P}(\mathrm{C})$ ?
Solution:

$
\begin{aligned}
& \mathrm{P}(\mathrm{B})=2 \mathrm{P}(\mathrm{A}) \\
& \mathrm{P}(\mathrm{C})=3 \mathrm{P}(\mathrm{A}) \\
& \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{4}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{8} \text {, } \\
& \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\frac{9}{10}, \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}) \frac{1}{15} \\
& \therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})- \\
& P(B \cap C)-P(A \cap C) \\
& +P(A \cap B \cap C) \\
& \frac{9}{10}=\mathrm{P}(\mathrm{A})+2 \mathrm{P}(\mathrm{A})+3(\mathrm{P}(\mathrm{A}))-\frac{1}{6}-\frac{1}{4}-\frac{1}{8}+\frac{1}{15} \\
& \frac{9}{10}=6 \mathrm{P}(\mathrm{A})+\left(\frac{-60-90-45+24}{360}\right) \\
& \frac{9}{10}=6 \mathrm{P}(\mathrm{A})+\frac{-171}{360} \\
& 6 \mathrm{P}(\mathrm{A})=\frac{9}{10}+\frac{171}{360}=\frac{324+171}{360}=\frac{495}{360} \\
& \mathrm{P}(\mathrm{A})=\frac{495}{360} \times \frac{1}{6}=\frac{11}{48} \\
& \therefore \mathrm{P}(\mathrm{B})=2 \mathrm{P}(\mathrm{A})=2 \times \frac{11}{48}=\frac{11}{24} \\
& \mathrm{P}(\mathrm{C})=3 \mathrm{P}(\mathrm{A})=3 \times \frac{11}{48}=\frac{11}{16} \\
& \mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B}), \mathrm{P}(\mathrm{C})=\frac{11}{48}, \frac{11}{24}, \frac{11}{16} \\
&
\end{aligned}
$
Question 14.
In a class of 35 , students are numbered from 1 to 35 . The ratio of boys to girls is $4: 3$. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with a prime roll number or a girl with a composite roll number or an even roll number.
Solution:

$
\begin{aligned}
& \mathrm{n}(\mathrm{S})=35 \\
& n(\mathrm{~S})=35 \\
& n(\mathrm{~B})=\frac{4}{7} \times 35=20 \\
& n(\mathrm{G})=\frac{3}{7} \times 35=15 \\
& \mathrm{~B}=\{1,2,3,4,5,6,7, \ldots 20\} \\
& \mathrm{G}=\{21,22, \ldots, 35\}
\end{aligned}
$
Boy with prime roll number $\rightarrow A$
$
\begin{aligned}
\mathrm{A} & =\{2,3,5,7,11,13,17,19\} \\
n(\mathrm{~A}) & =8 \\
\mathrm{P}(\mathrm{A}) & =\frac{8}{35}
\end{aligned}
$
Girl with composite roll number. $\rightarrow \mathrm{C}$
$
\begin{aligned}
\mathrm{C}= & \begin{array}{c}
\{21,22,24,25,26,27, \\
28,30,32,33,34,35\}
\end{array} \\
n(\mathrm{C})= & 12, \mathrm{P}(\mathrm{GC})=\frac{12}{35}
\end{aligned}
$

Student with even roll number - E
$
\begin{aligned}
& \mathrm{E}=\{2,4,6,8,10,12,14,16,18,20,22,24,26,28,30, \\
& n(\mathrm{E})=17, \mathrm{P}(\mathrm{E})=\frac{17}{35} \\
& \mathrm{~A} \cap \mathrm{C}=\{\}, n(\mathrm{~A} \cap \mathrm{C})=0, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{0}{35} \\
& \mathrm{C} \cap \mathrm{E}=\{22,24,26,28,30,32,34\}, n(\mathrm{C} \cap \mathrm{E})=7 \\
& \mathrm{P}(\mathrm{C} \cap \mathrm{E})=\frac{7}{35}, \\
& \mathrm{E} \cap \mathrm{A}=\{2\}, n(\mathrm{E} \cap \mathrm{A})=1, \mathrm{P}(\mathrm{E} \cap \mathrm{A})=\frac{1}{35}, \\
& \mathrm{P}(\mathrm{A} \cap \mathrm{C} \cap \mathrm{E})=0 \\
& \therefore \mathrm{P}(\mathrm{A} \cup \mathrm{C} \cup \mathrm{E})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{A} \cap \mathrm{C}) \\
& -\mathrm{P}(\mathrm{C} \cap \mathrm{E})-\mathrm{P}(\mathrm{E} \cap \mathrm{A})+\mathrm{P}(\mathrm{A} \cap \mathrm{C} \cap \mathrm{E}) \\
& =\frac{8}{35}+\frac{12}{35}+\frac{17}{35}-\frac{0}{35}-\frac{7}{35}-\frac{2}{35}+\frac{0}{35} \\
& =\frac{37-9}{35}=\frac{28}{35}=\frac{4}{5}
\end{aligned}
$