Probability Test

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Probability Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

If A and B are the two events of a sample space such that $$P(A\,\cup\,B)\, =\, \displaystyle \frac{5}{6}$$, $$P(A\,\cap\,B)\, =\,\displaystyle \frac{1}{3}$$, $$P(B)\, =\, \displaystyle \frac{1}{3}$$. Find P(A)

A
$$\displaystyle \frac{1}{6}$$

B
$$\displaystyle \frac{2}{6}$$

C
$$\displaystyle \frac{4}{6}$$

D
$$\displaystyle \frac{5}{6}$$

Solution

$$P(A\,\cup\,B)\, =\, \displaystyle \frac{5}{6}$$, $$P(A\,\cap\,B)\, =\,\displaystyle \frac{1}{3}$$, $$P(B)\, =\, \displaystyle \frac{1}{3}$$
$$P(A \cup B) = P(A) + P(B) - P(A\cap B)$$
$$\frac{5}{6} = P(A) + \dfrac{1}{3} - \dfrac{1}{3}$$

$$-P(A) = \dfrac{1}{3} - \dfrac{1}{3} - \dfrac{5}{6}$$

$$-P(A) = - \dfrac{5}{6}$$
$$P(A) = \dfrac{5}{6}$$
Question 2
1 / -0

If four positive integers are taken at random and multiplied together, then the probability that the last digit is 1, 3, 7 or 9 is

A
$$1/8$$

B
$$2/7$$

C
$$1/625$$

D
$$16/625$$

Solution

In any number, the last digit can be $$0,1,2,3,4,5,6,7,8,9$$.
We want that the last digit in the product is an odd digit other than $$5$$,
i.e., it is any one of the digits $$1,3,7,9$$. This means that the product is not divisible by $$2$$ or $$5$$.
The probability that a number is divisible by $$2$$ or $$5$$ is $$\displaystyle\frac { 6 }{ 10 } $$.
[$$\because $$ In that case the last digit can be one of $$0,2,4,5,6,8$$]
$$\therefore$$ The probability that the number is not divisible by $$2$$ or $$5=$$ $$\displaystyle1-\frac { 6 }{ 10 } -\frac { 4 }{ 10 } =\frac { 2 }{ 5 } $$
In order that the product is not divisible by $$2$$ or $$5$$,
none of tthe constituent numbers should be divisible by $$2$$ or $$5$$ and its probability $$\displaystyle={ \left( \frac { 2 }{ 5 } \right) }^{ 4 }=\frac { 16 }{ 625 } $$
Question 3
1 / -0

An investment consultant predicts that the odds against the price of a certain stock will go up during the next week are $$2:1$$ and the odds in favor of the price remaining the same are $$1:3$$. The probability that the price of the stock will go down during the next week, is

A
$$\displaystyle \frac { 4 }{ 12 } $$

B
$$\displaystyle \frac { 5 }{ 12 } $$

C
$$\displaystyle \frac { 7 }{ 12 } $$

D
none of these

Solution

Let $$A$$ denote the event 'stock price will go up', and $$B$$ be the event 'stock price will remain same'.
Then $$\displaystyle P\left( A \right) =\frac { 1 }{ 3 } $$ and $$\displaystyle P\left( B \right) =\frac { 1 }{ 4 } $$.
$$\therefore P($$ stock price will either go up or remain same $$)$$ is
$$\displaystyle P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) =\frac { 1 }{ 3 } +\frac { 1 }{ 4 } =\frac { 7 }{ 12 } $$
Hence probability that stock price will go down is given by
$$\displaystyle P\left( \overline { A } \cap \overline { B } \right) =1-P\left( A\cup B \right) =1-\frac { 7 }{ 12 } =\frac { 5 }{ 12 } $$
Question 4
1 / -0

A bag contains $$17$$ tickets numbered $$1$$ to $$17$$. A ticket is drawn and replaced, then one more ticket is drawn and replaced. Probability that first drawn number is even and second is odd, is

A
$$\displaystyle\frac { 81 }{ 289 } $$

B
$$\displaystyle\frac { 72 }{ 289 } $$

C
$$\displaystyle\frac { 64 }{ 289 } $$

D
none of these

Solution

Let $$A$$ be the event that first drawn number is even and $$B$$ be event that the second number drawn number is odd.
Total number of even numbers $$\displaystyle=\left[ \frac { 17 }{ 2 } \right] =8$$
$$\displaystyle\Rightarrow P\left( A \right) =\frac { 8 }{ 17 } $$ and $$\displaystyle P\left( B \right) =\frac { 9 }{ 17 } \Rightarrow P\left( A\cap B \right) =\frac { 72 }{ 289 } $$
Question 5
1 / -0

An $$MBM$$ applies for a job in two firms $$X$$ and $$Y$$. The probability of his being selected in firm $$X$$ is $$0.7$$ and being rejected at $$Y$$ is $$0.5$$. The probability of at least one of his applications being rejected is $$0.6$$. The probability that he will be selected in one of the firms, is

A
$$0.6$$

B
$$0.4$$

C
$$0.8$$

D
None of these

Solution

Let $$A$$ and $$B$$ denote the events that the person is selected in firms $$X$$ and $$Y$$ respectively.
Then in the usual notations, we are given:
$$P\left( A \right) =0.7\Rightarrow P\left( \overline { A } \right) =1-0.7=0.3\\ P\left( \overline { B } \right) =0.5\Rightarrow P\left( B \right) =1-0.5=0.5$$
and $$P\left( \overline { A } \cup \overline { B } \right) =0.6$$
The probability that the persons will be selected in one of the two firms $$X$$ or $$Y$$ is given by:
$$P\left( A\cup B \right) =1-P\left( \overline { A } \cap \overline { B } \right) =1-\left( P\left( \overline { A } \right) +P\left( \overline { B } \right) -P\left( \overline { A } \cup \overline { B } \right) \right) \\ =1-\left( 0.3+0.5-0.6 \right) =0.8$$
Question 6
1 / -0

If A, B and C are three events such that $$P(B)=\dfrac {3}{4}, P(A\cap B\cap C')=\dfrac {1}{3}$$ and $$P(A'\cap B\cap C')=\dfrac {1}{3}$$, then $$P(B\cap C)$$ is equal to

A
$$\dfrac {1}{12}$$

B
$$\dfrac {1}{6}$$

C
$$\dfrac {1}{15}$$

D
$$\dfrac {1}{9}$$

Solution

We have
$$P(B\cap C')=P[(A\cap A')\cap (B\cap C')]$$
$$\displaystyle =P(A\cap B\cap C')+P(A'\cap B\cap C') =\frac {1}{3}+\frac {1}{3}=\frac {2}{3}$$
Now, $$\displaystyle P(B\cap C)=P(B)-P(B\cap C') =\frac {3}{4}-\frac {2}{3}=\frac {1}{12}$$
Question 7
1 / -0

The probability that at least one of $$A$$ and $$B$$ occurs is $$0.6$$ and probability that they occur simultaneously is $$0.3$$, then $$P(A')+P(B')$$ is:

A
$$0.9$$

B
$$1.15$$

C
$$1.1$$

D
$$1.2$$

Solution

We have have $$P(A\cup B)=0.6$$ and $$P(A\cap B)=0.3$$.
We know that
$$P(A)+P(B)=P(A\cup B)+(A\cap B)=0.6+0.3=0.9$$
$$\therefore P(A')+P(B')=1-P(B)+1-P(A)=2-0.9=1.1$$
Question 8
1 / -0

What is the probability that a leap year has $$53$$ Sundays?

A
$$\displaystyle\frac{1}{7}$$

B
$$\displaystyle\frac{12}{52}$$

C
$$\displaystyle\frac{2}{7}$$

D
$$\displaystyle\frac{1}{52}$$

Solution

A leap year has $$366$$ days = $$ 52 \times 7 + 2 $$ days
That means there will be $$52$$ Sunday/Monday.../Saturdays plus $$2$$ additional days. Now these $$2 $$ additional days would be combination of any of the successive two days.
i)Sunday $$+$$ Monday
ii)Monday $$+$$ Tuesday
iii)Tuesday $$+$$ Wednesday
iv)Wednesday $$+$$ Thursday
v)Thursday $$+$$ Friday
vi)Friday $$+$$ Saturday
vii)Saturday $$+$$ Sunday
Now the probability of two Sundays will be (i) & (vii) $$ = 2$$ outcomes out of above $$7$$ outcomes. So the probability of having two additional Sundays
$$= 2/7.$$
Question 9
1 / -0

A die is thrown twice. What is the probability that
$$(i)$$ 3 will not come up either time?
$$(ii)$$ 6 will come up at least once?

A
$$(i) \displaystyle\frac{12}{25} \\ (ii) \displaystyle\frac{16}{25}$$

B
$$(i) \displaystyle\frac{18}{25} \\ (ii)\displaystyle\frac{9}{25}$$

C
$$(i) \displaystyle\frac{25}{36} \\ (ii) \displaystyle\frac{11}{36}$$

D
$$(i) \displaystyle\frac{23}{36} \\ (ii) \displaystyle\frac{17}{36}$$

Solution

When a dice is thrown twice, the possible 36 outcomes.
$$ 3 $$ will not come up either time, when pair of outcomes are anything except $$ (1,3), (3, 1), (3, 3), (2, 3), (3,2), (3,6), (6,3) (4,3), (3,4) , (3,5),(5,3) $$

Probability that $$ 3 $$ will not come up either time $$ = \dfrac {36-11}{36} = \dfrac {25}{36} $$

$$ 6 $$ will come up at least once, when pair of outcomes are $$ (1,6), (6,1), (6,2), (2,6), (3,6), (6,3), (4,6), (6,4), (6,5), (5,6), (6,6) $$

Probability that $$ 6 $$ will come up at least once $$ = \dfrac {11}{36} $$
Question 10
1 / -0

The probabilities of three events $$A,B$$ and $$C$$ are $$P\left( A \right) =0.6,P\left( B \right) =0.4,P\left( C \right) =0.5$$,also
$$P\left( A\cup B \right) =0.8,P\left( A\cap C \right) =0.3,P\left( A\cup B\cup C \right) \ge 0.85,P\left( A\cap B\cap C \right) =0.2$$
and $$P\left( B\cap C \right) ={ p }_{ 1 }$$. Then

A
$${ p }_{ 1 }\ge 0.35$$

B
$${ p }_{ 1 }\le 0.2$$

C
$$0.2\le { p }_{ 1 }\le 0.35$$

D
None of these

Solution

$$0.85\le P\left( A\cup B\cup C \right) \le 1\\ P\left( A\cap B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cup B \right) =0.2\\ \therefore P\left( A\cup B\cup C \right) =0.6+0.4+0.5-0.2-0.3-{ p }_{ 1 }+0.2\\ \Rightarrow { p }_{ 1 }=1.2-P\left( A\cup B\cup C \right) \\ \because 0.85\le P\left( A\cup B\cup C \right) \le 1\Rightarrow 0.2\le { p }_{ 1 }\le 0.35$$

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Probability Test - 32

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Probability Test - 32

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SHARING IS CARING

Question 1

$$\displaystyle \frac{1}{6}$$

$$\displaystyle \frac{2}{6}$$

$$\displaystyle \frac{4}{6}$$

$$\displaystyle \frac{5}{6}$$

Solution

Question 2

$$1/8$$

$$2/7$$

$$1/625$$

$$16/625$$

Solution

Question 3

$$\displaystyle \frac { 4 }{ 12 } $$

$$\displaystyle \frac { 5 }{ 12 } $$

$$\displaystyle \frac { 7 }{ 12 } $$

none of these

Solution

Question 4

$$\displaystyle\frac { 81 }{ 289 } $$

$$\displaystyle\frac { 72 }{ 289 } $$

$$\displaystyle\frac { 64 }{ 289 } $$

none of these

Solution

Question 5

$$0.6$$

$$0.4$$

$$0.8$$

None of these

Solution

Question 6

$$\dfrac {1}{12}$$

$$\dfrac {1}{6}$$

$$\dfrac {1}{15}$$

$$\dfrac {1}{9}$$

Solution

Question 7

$$0.9$$

$$1.15$$

$$1.1$$

$$1.2$$

Solution

Question 8

$$\displaystyle\frac{1}{7}$$

$$\displaystyle\frac{12}{52}$$

$$\displaystyle\frac{2}{7}$$

$$\displaystyle\frac{1}{52}$$

Solution

Question 9

$$(i) \displaystyle\frac{12}{25} \\ (ii) \displaystyle\frac{16}{25}$$

$$(i) \displaystyle\frac{18}{25} \\ (ii)\displaystyle\frac{9}{25}$$

$$(i) \displaystyle\frac{25}{36} \\ (ii) \displaystyle\frac{11}{36}$$

$$(i) \displaystyle\frac{23}{36} \\ (ii) \displaystyle\frac{17}{36}$$

Solution

Question 10

$${ p }_{ 1 }\ge 0.35$$

$${ p }_{ 1 }\le 0.2$$

$$0.2\le { p }_{ 1 }\le 0.35$$

None of these

Solution

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