Question

In four throws with a pair of dice what is the chance of throwing doublets at least twice?

Answer 1

Answer:

Chances of doublets = = 171/1296.

Step-by-step explanation:

When we throw a pair of dice, total chances are 6x6 = 36.

Options of doublets when we throw 2 dice at a time are 6. (11, 22,33,44,55,66)

So probability of doublets in 2 dice throw is 6/36 = 1/6. = p

Probability of no doublets = 1 – p = 5/6 = q.

Let P(x) is the event, chances of event occurring x times.  

P(x) is given as nCr * p^r * q^(n-r)

n = 4.

P(At least 2 doublets in 4 throws) = NOT( 0 doublets + 1 doublet)

= 1 – (P(0) + P(1))

= 1 – (4C0 * (1/6)^0 * (5/6)^4  +  4C1 * (1/6)^1 * (5/6)^3)

= 1 – ((5/6)^4  + 4*(1/6)*(5/6)^3)

= 171/1296.

Answer 2

Answer:

4C2 X (6/36)^2 X (30/36)^2 = 6 x (1/36) x (25/36) = 25/216

Step-by-step explanation: