1A number is selected at random from the first 50 positive integers. What is the probability that it is a Prime number? A3/20 B5/20 C4/20 D7/20
Answer: A
Explanation:
There are 15 Prime numbers from 1 to 50.
$\therefore$ Required probability $ = \dfrac{{15}}{{100}} = \dfrac{3}{{20}}$
2Four Fair coins are tossed together. Find the probability that tail appears on exactly one of the coins. A2/4 B3/4 C1/4 D2/5
Answer: C
Explanation:
Total number of cases when $n$ coins are tossed $=2^n$
The total number of cases when 4 coins are tossed $= 2^4\; = 16$
As Tail has to appear on exactly one of the coins, the favourable cases $= THHH, HTHH, HHTH, HHHT$.
$\therefore$ Required probability $ = \dfrac{4}{{16}} = \dfrac{1}{4}$
Alternative Method:
Number of way of getting exactly $r$ tails or heads on tossing $n$ coins = $\dfrac{{{}^n{C_r}}}{{{2^n}}}$
$\therefore$ $\dfrac{{{}^4{C_1}}}{{{2^4}}} = \dfrac{4}{{16}} = \dfrac{1}{4}$
3 Four Fair coins are tossed together. Find the probability that tail appears on exactly two of the coins. A7/8 B3/4 C5/8 D3/8
Answer: D
Explanation:
Total number of cases when $n$ coins are tossed $=2^n$
The total number of cases $= 2^4\; = 16$
Tail has to appear on exactly 2 coins, i.e $TTHH$ in some order.
Total ways of arranging two $H$'s and two $T$'s = $\dfrac{{4!}}{{2! \times 2!}}$ $= \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 2}} = 6$
$\bigl($Formula:Number of ways of arranging $n$ items of which $p$ are alike, $q$ alike and so on $ = \dfrac{{n!}}{{p! \times q!}}\Biggr)$
$\therefore$Required probability $ = \dfrac{6}{{16}} = \dfrac{3}{8}$
Alternative Method:
Number of way of getting exactly $r$ tails or heads on tossing $n$ coins = $\dfrac{{{}^n{C_r}}}{{{2^n}}}$
$\therefore$ $\dfrac{{{}^4{C_2}}}{{{2^4}}} = \dfrac{6}{{16}} = \dfrac{3}{8}$
4Eight unbiased coins are tossed simultaneously. What is the probability that at least 2 tails turn up? A233/256 B240/256 C222/256 D247/256
Answer: D
Explanation:
Number of way of getting exactly $r$ tails or heads on tossing $n$ coins = $\dfrac{{{}^n{C_r}}}{{{2^n}}}$
Probability of getting atleast 2 tails = 1 - (probability of getting no tails + probability of getting exactly one tail)
$\therefore$ Probability that at least two tails turn up $ = 1 - \left( {\dfrac{{{}^8{C_0} + {}^8{C_1}}}{{{2^8}}}} \right)$ $=1 - \left( {\dfrac{{1 + 8}}{{256}}} \right) = 1 - \left( {\dfrac{9}{{256}}} \right)$ $={\dfrac{{247}}{{256}}}$
5An unbiased dice is rolled. What is the probability that the number appearing on the dice is even? A1/3 B1/6 C1/2 D1/4
Answer: A
Explanation:
When a dice is rolled total outcomes $= 6$
Number of favourable outcomes $= 3(i.e., 2,4,6)$
$\therefore$Required probability $ = \dfrac{2}{6} = \dfrac{1}{3}$
6When a fair cubical dice is rolled. What is the probability of getting a Prime number? A1/3 B1/4 C1/2 D2/3
Answer: C
Explanation:
On rolling a fair die, the total possibilities are 6.
The number of ways of getting a Prime number $= 3(i.e., 2,3,5)$
$\therefore$The required probability = $ = \dfrac{3}{6} = \dfrac{1}{2}$
