Probability Test

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Probability Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

The contents of urn I and II are as follows:
Urn I: 4 white and 5 black balls
Urn II: 3 white and 6 black balls
One urn is chosen at random and a ball is drawn and its colour is noted and replaced back to the urn. Again a ball is drawn from the same urn colour is noted and replaced. The process is repeated 4 times and as a result one ball of white colour and 3 of black colour are noted. Find the probability the chosen urn was I.

A
$$\displaystyle \frac{125}{287}$$

B
$$\displaystyle \frac{64}{127}$$

C
$$\displaystyle \frac{25}{287}$$

D
$$\displaystyle \frac{79}{192}$$

Solution
Question 2
1 / -0

The ratio of the number of trucks along a highway, on which a petrol pump is located, to the number of cars running along the same highway is 3 : 2. It is known that an average of one truck in thirty trucks and two cars in fifty cars stop at the petrol pump to be filled up with the fuel. If a vehicle stops at the petrol pump to be filled up with the fuel, find the probability that it is a car

A
$$\displaystyle \frac {4}{9}$$

B
$$\displaystyle \frac {9}{250}$$

C
$$\displaystyle \frac {3}{5}$$

D
$$\displaystyle \frac {1}{30}$$

Solution

C : Vehicle is Car $$P(\overline C ) = 2/5$$
$$\overline C$$ : Vehicle is Truck $$P(C) = 3/5$$
E : Vehicle stops for fuel $$P(E/C) = 2/50$$
$$P(\overline E /\overline C) = 1 / 30$$
$$P(E) = P(CE \cup C\overline E)$$
$$= P(CE) + P(\overline C E)$$
$$\displaystyle =\frac {2}{5}\left (\frac {2}{50}\right )+\frac {3}{5}.\frac {1}{30}=\frac {1}{10}\left [\frac {4}{25}+\frac {1}{5}\right ]=\frac {9}{250}$$
$$\displaystyle P(C/E)=\frac {P(CE)}{P(E)}=\frac {4/250}{9/250}=\frac {4}{9}$$
Question 3
1 / -0

In a series of 3 independent trials, the probability of exactly 2 success is 12 times as large as the probability of 3 successes. The probability of a success in each trail is

A
$$\dfrac{1}{5}$$

B
$$\dfrac{2}{5}$$

C
$$\dfrac{3}{5}$$

D
$$\dfrac{4}{5}$$

Solution

For a binomial experiment consisting of n trials, the probability of exactly
k successes is
$$P(k \text{ successes}) = ^nC_kp^k(1-p)^{n-k}$$
where the probability of success on each trial is p.
According to the question,
$$n=3,, P(\text{2 successes})=12\times P(\text{3 successes})$$
To find:
The probability of a success in each trail, p
$$P(\text{2 successes})=12\times P(\text{3 successes})\\\implies ^3C_2p^2(1-p)^{3-2}=12 \times ^3C_3p^3(1-p)^{3-3}\\\implies 3\times p^2(1-p)^1=12\times 1\times p^3(1-p)^0\\\implies 3p^2-3p^3=12p^3$$
Divide throughout by $$3p^2$$, we get
$$1-p=4p\\\implies 5p=1\\\implies p=\dfrac 15$$
Question 4
1 / -0

A signal which can be green or red with probability $$\displaystyle \frac{4}{5}$$ and $$\displaystyle \frac{1}{5}$$, respectively, is received at station A and then transmitted to station B. The probability of each station receiving the signal correctly is $$\displaystyle \frac{3}{4}$$. If the signal received at station B is green, then the probability that the original signal was green is

A
$$\displaystyle \frac{3}{5}$$

B
$$\displaystyle \frac{6}{7}$$

C
$$\displaystyle \frac{20}{23}$$

D
$$\displaystyle \frac{9}{20}$$

Solution

Event $$G$$ = original signal is green
$$E_1=A$$ receives the signal correct
$$E_2=B$$ receives the signal correct
E = signal received by B is green
$$P(\text{signal received by B is green}) = P(GE_1E_2)+ P(G\cap {E_1}\cap {E_2})+ P(\cap GE_1\cap{E_2})+ P(\cap G\cap {E_1}E_2)$$
$$P(E)=\dfrac {46}{5\times 16}$$
$$ P(G/E)=\dfrac {\dfrac {40}5\times 16}{\dfrac {46}5\times16}=\dfrac {20}{23}.$$
Question 5
1 / -0

There are two urns. There are $$m$$ white & $$n$$ black balls in the first urn and $$p$$ white & $$q$$ black balls in the second urn. One ball is taken from the first urn & placed into the second. Now, the probability of drawing a white ball from the second urn is

A
$$\displaystyle \frac{pm+(p+1)n}{(m+n)(p+q+1)}$$

B
$$\displaystyle \frac {(p+1)m+pn}{(m+n)(p+q+1)}$$

C
$$\displaystyle \frac{qm+(q+1)n}{(m+n)(p+q+1)}$$

D
$$\displaystyle \frac {(q+1)m+qn}{(m+n)(p+q+1)}$$

Solution

Given:
Urn 1: $$U_1$$ has m white and n black balls.
Urn 2: $$U_2$$ has p white and q black balls.
The probability of white ball in urn 1 is $$\dfrac m{m+n}$$ and of black ball is $$\dfrac n{m+n}$$
Similarly, the probability of white ball in urn 2 is $$\dfrac p{p+q}$$ and of a black ball is $$\dfrac q{p+q}$$
One ball is taken from the first urn & placed into the second.
Now,
Case (i): If the transferred ball is white,
Urn 2: $$U_2$$ has (p+1) white and q black balls.
The probability of white ball in urn 2 is $$\dfrac {p+1}{p+q+1}$$ and of a black ball is $$\dfrac q{p+q+1}$$
Case (ii): If the transferred ball is black,
Urn 2: $$U_2$$ has p white and (q+1) black balls.
The probability of white ball in urn 2 is $$\dfrac p{p+q+1}$$ and of a black ball is $$\dfrac {q+1}{p+q+1}$$
Hence, the probability of drawing a white ball from the second urn is
$$\dfrac {m}{m+n}\times \dfrac {p+1}{p+q+1}+\dfrac {n}{m+n}\times \dfrac {p}{p+q+1}\\\implies \dfrac {m(p+1)+np}{(m+n)(p+q+1)}$$
Question 6
1 / -0

One ticket is selected at random from $$100$$ tickets numbered $$100,01, 02, .... 98, 99.$$ If $$x_{1}$$ and $$x_{2} $$ denotes the sum and product of the digits on the tickets, then $$ P((x_{1} =9)/(x_{2} =0))$$ is equal to

A
$$2/19$$

B
$$19/100$$

C
$$1/50$$

D
none of these

Solution

Total number of ways , $$n(S)=100$$
Sample space for product of digits equal to zero is = $${01 , 02 ,03 ,04 ,05 , 06 ,07 , 08 , 09 , 10 , 20 ,30 , 40 ,50 , 60 ,70 ,80 , 90,100}$$
So,
$$n(x_2)=19$$
$$P(x_2)=\dfrac {19}{100}$$
Sample space for sum of digits equal to 9 is =$${09, 18 , 27 , 36 , 45 , 54 , 63 , 72 , 81 , 90}$$
Now, sample space for the sum and product of the digits on the tickets is =$${09,90} $$
Now $$P(x_1\cap x_2)=\dfrac {2}{100}$$
Hence, $$P\left(\dfrac {(x_1=9)}{(x_2=0)}\right)=\dfrac {P(x_1\cap x_2)}{P(x_2)}=\dfrac {\dfrac 2{100}}{\dfrac {19}{100}}=\dfrac 2{19}$$
Question 7
1 / -0

A fair coin is tossed five times. If the out comes are $$2$$ heads and $$3$$ tails (in some order), then what is the probability that the fourth toss is a head?

A
$$\displaystyle \frac{1}{4}$$

B
$$\displaystyle \frac{2}{5}$$

C
$$\displaystyle \frac{1}{2}$$

D
$$\displaystyle \frac{3}{5}$$

Solution

A fair coin is tossed $$5$$ times
$$\Rightarrow E=2$$ heads are obtained on first two tosses.
$$\Rightarrow F=3$$ tails are obtained on next three tosses.
Sample space,
$$\Rightarrow S=[HHHHH,HHHHT,HHHTH,HHTHH,HTHHH,THHHH,TTHHH,HTTHH,$$
$$HHTTH,HHHTT,THHHT,HTHTH,THTHH,HHTHT,TTTHH,HTTTH,HHTTT,$$
$$THHTT,TTHHT,THTHT,HTHTT,TTHTH,TTTTH,TTTHT,TTHTT,THTTT,HTTTT,$$
$$TTTTT]$$
$$\Rightarrow$$ Sample space $$=28$$
$$\Rightarrow E=2$$ heads at first two tosses $$=\dfrac{8}{28}$$
$$\Rightarrow F=3$$ tails on next three tosses $$=\dfrac{4}{28}.$$
$$\Rightarrow P\left( \dfrac { E }{ F } \right) =\dfrac{n(E\cap F)}{n(F)}$$
$$\Rightarrow P\left( \dfrac { E }{ F } \right)=\dfrac{\dfrac{8}{28}}{\dfrac{4}{28}}$$
$$=2.$$
$$\Rightarrow n(F)=\dfrac{P\left( \dfrac { E }{ F } \right)}{no.\;of\;tosses}$$
$$=\dfrac{2}{5}.$$
Hence, the answer is $$\dfrac{2}{5}.$$
Question 8
1 / -0

A boy has 20% chance of hitting at a target. Let p denote the probability of hitting the target for the first time at the nth trial. If $$p$$ satisfies the inequality $$ 625p^{2} - 175p + 12 < 0$$ then value of $$n$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Let p(n)=Hitting for the 1st time at the n-th trial
$$\implies p(n)=0.8^{n-1}\times0.2$$
The boundary of the inequality $$62p^2-175p+12=0$$
is the solution of a quadratic equation in p :
or, $$175^2-4\times12\times625=625=25^2$$
$$\implies p=\dfrac {175\pm25}{1250}=\dfrac 3{25}$$ or $$\dfrac 4{25}$$
So p(n) is negative between those two values.
$$p(n)=\dfrac 3{25}=0.8^{n-1}\times 0.2\\\implies \dfrac 35=0.8^{n-1}\\\implies \log\left(\dfrac 35\right)=(n-1)\log (0.8)\\\implies n=1+\dfrac {\log \left(\dfrac 35\right)}{\log (0.8)}=3.289$$
$$p(n)=\dfrac 4{25}=0.8^{n-1}\times 0.2\\\implies \dfrac 45=0.8^{n-1}\\\implies \log\left(\dfrac 45\right)=(n-1)\log (0.8)\\\implies n=1+\dfrac {\log \left(\dfrac 45\right)}{\log (0.8)}=2$$
Hence, $$2<n<3.289\implies n=3$$
Question 9
1 / -0

When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8.

A
$$\dfrac{1}{5}$$

B
$$\dfrac{2}{5}$$

C
$$\dfrac{3}{5}$$

D
None of these

Solution

Let the event of getting a greater number on the first die be $$G$$.
There are $$5$$ ways to get a sum of $$8$$ when two dice are rolled $$=\{(2,6),(3,5),(4,4), (5,3),(6,2)\}$$
And there are two ways where the number on the first die is greater than the one on the second given that the sum should equal $$8$$, $$G = \{(5,3), (6,2)\}$$
Therefore, $$P($$Sum equals $$8)$$ $$=\dfrac{5}{36}$$ and $$P(G)$$ $$=\dfrac{2}{36}$$
Now, $$P(G\ |$$ Sum equals $$8)$$ $$= \dfrac{P(G \,and \,sum \,equals\, 8)}{P(sum \,equals\, 8)}$$
$$= \dfrac{(2/36)}{(5/36)}$$ $$=\dfrac{2}{5}$$
Question 10
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In a set of $$10$$ coins, $$2$$ coins are with heads on both the sides. A coin is selected at random from this set and tossed five times. If all the five times, the result was heads, find the probability that the selected coin had heads on both the sides.

A
$$\dfrac16$$

B
$$\dfrac89$$

C
$$\dfrac23$$

D
$$\dfrac2{10}$$

Solution

Let $$E_1, E_2, A$$ be the events defined as follows :
$$E_1$$= selecting a coin having a head on both the sides.
$$E_2$$= selecting a coin not having a head on both the sides.
A = Getting all heads when a coin is tossed five times.
To find:
The probability that the selected coin had heads on both the sides: $$P\left(\dfrac {E_1}A\right)$$
There are 2 coins having heads on both the sides
$$P(E_1)=\dfrac {^2C_1}{^{10}C_1}=\dfrac 2{10}$$
There are 8 coins not having heads on both the sides
$$P(E_2)=\dfrac {^8C_1}{^{10}C_1}=\dfrac 8{10}$$
$$P\left(\dfrac A{E_1}\right)=1^5=1\\\implies P\left(\dfrac A{E_2}\right)=\left(\dfrac 12\right)^5$$
By Bayee's Theorem, we have
$$P\left(\dfrac {E_1}A\right)=\dfrac {P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}=\dfrac {\dfrac 2{10}\times 1}{\dfrac 2{10}\times 1+\dfrac 8{10}\times \left(\dfrac 12\right)^5}\\\implies P\left(\dfrac {E_1}A\right)=\dfrac 2{2+\dfrac 8{32}}=\dfrac 89$$

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Re-Attempt Weekly Quiz Competition

Probability Test - 51

Result

Probability Test - 51

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SHARING IS CARING

Question 1

$$\displaystyle \frac{125}{287}$$

$$\displaystyle \frac{64}{127}$$

$$\displaystyle \frac{25}{287}$$

$$\displaystyle \frac{79}{192}$$

Solution

Question 2

$$\displaystyle \frac {4}{9}$$

$$\displaystyle \frac {9}{250}$$

$$\displaystyle \frac {3}{5}$$

$$\displaystyle \frac {1}{30}$$

Solution

Question 3

$$\dfrac{1}{5}$$

$$\dfrac{2}{5}$$

$$\dfrac{3}{5}$$

$$\dfrac{4}{5}$$

Solution

Question 4

$$\displaystyle \frac{3}{5}$$

$$\displaystyle \frac{6}{7}$$

$$\displaystyle \frac{20}{23}$$

$$\displaystyle \frac{9}{20}$$

Solution

Question 5

$$\displaystyle \frac{pm+(p+1)n}{(m+n)(p+q+1)}$$

$$\displaystyle \frac {(p+1)m+pn}{(m+n)(p+q+1)}$$

$$\displaystyle \frac{qm+(q+1)n}{(m+n)(p+q+1)}$$

$$\displaystyle \frac {(q+1)m+qn}{(m+n)(p+q+1)}$$

Solution

Question 6

$$2/19$$

$$19/100$$

$$1/50$$

none of these

Solution

Question 7

$$\displaystyle \frac{1}{4}$$

$$\displaystyle \frac{2}{5}$$

$$\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{3}{5}$$

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 9

$$\dfrac{1}{5}$$

$$\dfrac{2}{5}$$

$$\dfrac{3}{5}$$

None of these

Solution

Question 10

$$\dfrac16$$

$$\dfrac89$$

$$\dfrac23$$

$$\dfrac2{10}$$

Solution

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