Question

A coin and a die are thrown simultaneously. Find the probability of getting i) tail on the coin and 6 on the die ii) head on the coin and even number in the die.

Please help!​

Answer 1

• i.
  `\boxed{(1)/(12)}`
• ii.
  `\boxed{(1)/(4)}`

Explanation:

Finding the probability of getting

i) tail on the coin and 6 on the die

Solution:

Favorable cases of getting tail in coin=1

Total cases in coin=2

Favorable cases of getting 6 in dice=1

Total cases in dice=6

`\tt \:Probability \:of \:\:getting \:\:tail\:\: in\:\: coin\:\:P(T)= (favourable \:\:cases )/(Total \:\:no.\:\:cases)=(1)/(2)`

`\tt \:Probability \:of \:\:getting \:\:6\:\: in\:\: dice\:P(6): (favourable \:\:cases )/(Total \:\:no.\:\:cases)=(1)/(6)`

Now,

Probability of getting tail on the coin and 6 on the dice = P(T)*P(6)

=
`\tt (1)/(2)*(1)/(6)`

=
`\tt (1)/(12)`

Therefore answer is
`\tt (1)/(12)`

`\hrulefill`

ii) head on the coin and even number in the die

Solution:

Favorable cases of getting head in coin=1

Total cases in coin=2

Favorable cases of getting even number in dice=3

Total cases in dice=6

`\tt \:Probability \:of \:\:getting \:\:tail\:\: in\:\: coin\:\:P(H)= (favourable \:\:cases )/(Total \:\:no.\:\:cases)=(1)/(2)`

`\tt \:Probability \:of \:\:getting \:\:even\:\:no.\:\: in\:\: dice\:P(E): (favourable \:\:cases )/(Total \:\:no.\:\:cases)=(3)/(6)=(1)/(2)`

Now,

Probability of getting tail on the coin and 6 on the dice = P(H)*P(E)

=
`\tt (1)/(2)*(1)/(2)`

=
`\tt (1)/(4)`

Therefore answer is
`\tt (1)/(4)`