PROBABILITY QUESTIONS

subhamalya · Joined: 05 Jan 2005 Posts: 10

Hi! plz help me solve these 3 sums on probability

In 4 throws with a pair of dice, what is the chance of throwing a double twice?
1. 11/216 2. 25/216 3. 35/126 4. 45/216

3 of the 6 vertices of a regular hexagon are chosen at random.The
probability that the triangle with these vertices is equilateral is
1. 1/10 2. 3/10 3. 1/5 4. 4/10

7 white balls & 3 black balls are randomly placed in a row. Find the probability that no two black balls are placed adjacent to each other.
1.7/15 2. 2/15 3. 3/7 4.2/7

manish4iim · Joined: 04 Jan 2005 Posts: 11

The total ways in which the 10 balls can be arranged is 10!. As it is not explicity stated, it is to be taken that the balls are not identical.

This will be denominator

Now, the 7 white balls can be arranged in 7! ways. There will be 8 vacant slots available amongst the 7 white balls.

_ w _ w _ w _ w _ w _ w _ w _

The three black balls can be placed amongst the 8 in 8P3 ways.

Therefore, required probability is (8P3 * 7!) / 10!

= (8 * 7 * 6 * 7!) / 10 * 9 * 8 * 7!

= 42/90 = 7/15

Choice (1)

shaji · Posted: Thu May 26, 2005 1:26 pm Post subject: Re: PROBABILITY QUESTIONS

Problem1.

Use the binomial theorem:

4C2*(1/6)^2*(5/6)^2=25/216.

The other problems have solutions on the forum.

Shaji.

neerajj · Joined: 29 Aug 2005 Posts: 7

Hi Guys,

The first problem is not clear .. "what is the chance of throwing a double twice".

The second problem will have probability as 1/10 as ,

Total of number of triangles possible using the vertices of hexagon is 6C3 = 20
Total number of equilateral triangles possible = 2 .

Let us choose amongst one of the 6 vertices as one vertex of the triangle, now the other vertex can be one vertex away from this vertex in either clockwise and anticlockwise direction . All the other 3 vertex will already be part of this triangle.