M.C.Q (1 Marks)

MCQ 1511 Mark

A die is thrown then find the probability of getting a perfect square.

✓
$\frac{1}{3}$
B
$\frac{1}{2}$
C
$\frac{2}{3}$
D
$0$

Answer

Correct option: A.

$\frac{1}{3}$

Sample space $= 1, 2, 3, 4, 5, 6$
A perfect square in samples $=1, 4 = 2$
Probability of getting perfect square $=\frac{1}{3}$

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MCQ 1521 Mark

The set of all possible outcomes of any experiment is called:

A
Event
B
Random experiment
✓
Sample space
D
Sample point

Answer

Correct option: C.

Sample space

The set of all possible outcomes of any experiment is called sample space.
Example: toss a coin,
Sample space, $S = \{H, T\}$
$H =$ head
$T =$ tail

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MCQ 1531 Mark

The sample space for choosing $2$ letters at random from a set of $55$ vowels is $..........$

A
$\{e, i, p\}$
✓
$\{n, o, u\}$
C
$\{a, e, i, o, u\}$
D
$\{a, h\}$

Answer

Correct option: B.

$\{n, o, u\}$

Sample space is the collection of all possible outcomes.
So, sample space, $S = \{a, e, i, o, u\}$

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MCQ 1541 Mark

The radius of the circle passing through the point $(6, 2)$ and two of whose diameters are $x + y = 6$ and $x+2y = 4$ is:

A
$4$
B
$6$
C
$20$
✓
$\sqrt{20}$

Answer

Correct option: D.

$\sqrt{20}$

int of intersection of the given diameters is $(8, -2)$ which is the centre of
the circle Also the circle pass through the point $(6, 2),$
so the radius is
$=\sqrt{(8-6)^2+(-2-2)^2}$
$=\sqrt{20}$

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MCQ 1551 Mark

Choose the correct answer. If $M$ and $N$ are any two events, the probability that at least one of them occurs is:

A
$\text{P(M)}+\text{P(N)}-2\text{P(M}\cap\text{N)}$
✓
$\text{P(M)}+\text{P(N)}-\text{P(M}\cap\text{N)}$
C
$\text{P(M)}+\text{P(N)}+\text{P(M}\cap\text{N)}$
D
$\text{P(M)}+\text{P(N)}+2\text{P(M}\cap\text{N)}$

Answer

Correct option: B.

$\text{P(M)}+\text{P(N)}-\text{P(M}\cap\text{N)}$

If $M$ and $N$ are any two events.
$\text{P(M}\cup\text{N)}=\text{P(M)}+\text{P(N)}-\text{P(M}\cap\text{N)}$

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MCQ 1561 Mark

Two dice are thrown simultaneously. Find the probability of getting an even number as the sum.

A
$\frac{1}{6}$
✓
$\frac{1}{2}$
C
$\frac{5}{6}$
D
$\frac{1}{3}$

Answer

Correct option: B.

$\frac{1}{2}$

Total number of possible cases $= 36$
Favourable cases of getting even number as the sum
$= \{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), \\ (3, 5), (4, 2), (4, 4), (4, 6), (5,1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\}$
Total number of favourable cases $= 18$
$P($getting even number as the sum$) =\frac{18}{36}=\frac{1}{2}$

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MCQ 1571 Mark

One card is drawn from a pack of $52$ cards.The probability of getting a $10$ of black suit is:

✓
$\frac{1}{26}$
B
$\frac{1}{13}$
C
$\frac{3}{26}$
D
$\text{None}$

Answer

Correct option: A.

$\frac{1}{26}$

Favourable number of outcomes, with $10$ of black suit $= 2$
Total number of outcomes $= 52$
Thus, probabilit
$=\frac{2}{52}$
$=\frac{1}{26}$

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MATHS M.C.Q (1 Marks)