Question

Can someone answer this

Answer 1

Answer:

1/2

1/2

1/2

1/6

0

2/12

8/12

4/12

6/12

4/12

6/12

2/12

3/12,2/12

1

6/12

6/12

4/12

1/6

3/6

jo space khali chode hain unka matlab hai ki nhi aa rha hai

follow me and mark as brainlest

Answer 2

The sample space is,

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Total no. of outcomes = 6 × 6 = 36

Probability of getting an even number on first dice,

$\displaystyle\longrightarrow\sf{P(1)=\dfrac{18}{36}}=\bf{\dfrac{1}{2}}$

Probability of getting an odd number on first dice,

$\displaystyle\longrightarrow\sf{P(2)=\dfrac{18}{36}}=\bf{\dfrac{1}{2}}$

Probability of getting an even number as the sum,

$\displaystyle\longrightarrow\sf{P(3)=1-\dfrac{18}{36}}=\bf{\dfrac{1}{2}}$

Probability of getting a multiple of 5 as the sum,

$\displaystyle\longrightarrow\sf{P(4)}=\bf{\dfrac{7}{36}}$

Probability of getting a multiple of 7 as the sum,

$\displaystyle\longrightarrow\sf{P(5)=\dfrac{6}{36}}=\bf{\dfrac{1}{6}}$

Probability of getting a multiple of 3 as the sum,

$\displaystyle\longrightarrow\sf{P(6)=\dfrac{12}{36}}=\bf{\dfrac{1}{3}}$

Probability of getting a sum more than 7,

$\displaystyle\longrightarrow\sf{P(7)=\dfrac{15}{36}}=\bf{\dfrac{5}{12}}$

Probability of getting a sum greater than 9,

$\displaystyle\longrightarrow\sf{P(8)=\dfrac{6}{36}}=\bf{\dfrac{1}{6}}$

Probability of getting neither 9 nor 11 as the sum,

$\displaystyle\longrightarrow\sf{P(9)=1-\dfrac{4}{36}-\dfrac{2}{36}}=\bf{\dfrac{5}{6}}$

Probability of getting a sum less than 6,

$\displaystyle\longrightarrow\sf{P(10)=\dfrac{10}{36}}=\bf{\dfrac{5}{18}}$

Probability of getting a sum less than 7,

$\displaystyle\longrightarrow\sf{P(11)=\dfrac{5}{18}+\dfrac{5}{36}}=\bf{\dfrac{5}{12}}$

Probability of getting a multiple of 3 on one dice (except getting a multiple of 3 on both dices),

$\displaystyle\longrightarrow\sf{P(12)=\dfrac{16}{36}}=\bf{\dfrac{4}{9}}$

Probability of getting a multiple of 2 on one dice (except getting a multiple of 2 on both dices),

$\displaystyle\longrightarrow\sf{P(13)=\dfrac{18}{36}}=\bf{\dfrac{1}{2}}$

Probability of getting a multiple of 5 on one dice (except getting a multiple of 5 on both dices),

$\displaystyle\longrightarrow\sf{P(14)=\dfrac{10}{36}}=\bf{\dfrac{5}{18}}$

Probability of getting a multiple of 2 on one dice and a multiple of 3 on the other,

$\displaystyle\longrightarrow\sf{P(15)}=\bf{\dfrac{11}{36}}$

Probability of getting a doublet,

$\displaystyle\longrightarrow\sf{P(16)=\dfrac{6}{36}}=\bf{\dfrac{1}{6}}$

Probability of getting a doublet of even number,

$\displaystyle\longrightarrow\sf{P(17)=\dfrac{3}{36}}=\bf{\dfrac{1}{12}}$

Probability of getting a doublet of odd number,

$\displaystyle\longrightarrow\sf{P(18)=\dfrac{3}{36}}=\bf{\dfrac{1}{12}}$

Probability of getting a doublet of prime number,

$\displaystyle\longrightarrow\sf{P(19)=\dfrac{3}{36}}=\bf{\dfrac{1}{12}}$

Probability of getting a number other than 5 on any dice,

$\displaystyle\longrightarrow\sf{P(20)=1-\dfrac{11}{36}}=\bf{\dfrac{25}{36}}$

Probability of getting a number other than 3 on any dice,

$\displaystyle\longrightarrow\sf{P(21)=1-\dfrac{11}{36}}=\bf{\dfrac{25}{36}}$

Probability of getting a sum equal to 12,

$\displaystyle\longrightarrow\sf{P(22)}=\bf{\dfrac{1}{36}}$

Probability of getting a sum greater than or equal to 10,

$\displaystyle\longrightarrow\sf{P(23)=\dfrac{6}{36}}=\bf{\dfrac{1}{6}}$

Probability of getting a sum less than or equal to 10,

$\displaystyle\longrightarrow\sf{P(24)=1-\dfrac{3}{36}}=\bf{\dfrac{11}{12}}$

Probability of getting a prime number as the sum,

$\displaystyle\longrightarrow\sf{P(25)=\dfrac{15}{36}}=\bf{\dfrac{5}{12}}$

Probability of getting a product 6,

$\displaystyle\longrightarrow\sf{P(26)=\dfrac{4}{36}}=\bf{\dfrac{1}{9}}$

Probability of getting a product 12,

$\displaystyle\longrightarrow\sf{P(27)=\dfrac{4}{36}}=\bf{\dfrac{1}{9}}$

Probability of getting a product 7,

$\displaystyle\longrightarrow\sf{P(28)=\dfrac{2}{36}}=\bf{\dfrac{1}{18}}$

Probability of getting a product less than 9,

$\displaystyle\longrightarrow\sf{P(29)=1-\dfrac{1}{36}-\dfrac{2}{36}-\dfrac{4}{36}}=\bf{\dfrac{29}{36}}$