Question

A dice and a coin is tossed together find the probability that (1) a head and even number (2) a tail and a prime number​

Answer 1

Probability = \bold{\dfrac{1}{4}}

4

1

\bold{\large{\underline{\underline{\sf{StEp\:by\:stEp\:explanation:}}}}}

StEpbystEpexplanation:

GiVeN :

A die is rolled and a coin is tossed simultaneously

To FiNd :

The probability that the die shows an odd number and the coin shows head.

SoLuTiOn :

Let's first considere the case of coin.

Sample space for coin :

Let the sample space be S.

S = {H, T}

•°• n (S) = 2.

Let's now consider the die.

Sample space for die :

Sample space,

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

•°• n (S) = 6

Now, let's consider the case when the coin is tossed and a die is thrown simultaneously.

Sample space,

\bold{\sf{S=(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5)(T,6)}}S=(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5)(T,6)

•°• n (S) = 12

Now let's find the probability.

Let O be the event when the die shows an odd number and the coin shows head.

From the above sample space, we have head with 1,2,3,4,5 and 6.

Out of which odd numbers are :

1

3

5

•°• We infer that there are 3 possibilities of die showing an odd number and the coin showing head simultaneously.

Probability :

P (O) = \bold{\dfrac{n(O)}{n(S)}}

n(S)

n(O)

P (O) = \bold{\dfrac{3}{12}}

12

3

P (O) = \bold{\dfrac{1}{4}}

4

1

•°• Probability of die showing an odd number and the coin showing head is ¼