Four Coins Are Thrown Simultaneously From A Bridge. Coin A Is Thrown Vertically

When 4 coins are tossed simultaneously, what is the probability of getting 2 heads, at least 2 heads, and at most 2 heads?

When tossing four “fair” coins there are 16 equally likely outcomes, this can be found by taking the number of outcomes per event, 2; heads and tails, and raising it to the power of the number of events, 4. 2^4 = 16.There are 6 ways to get two heads; HHTT, HTHT, HTTH, THHT, THTH, TTHH. Mathematically this is 4C2 which is equal to (4!)/[(2!)(4–2)!] =[4x3x2x1]/[(2x1)(2x1)] = 24/4 = 6A) The probability of 2 heads is 6/16 = 3/8 = 37.5%.To find at least two heads, you will need to add on the probability of 3 heads and also 4 heads. to your answer for exactly two heads.The probability of 3 heads is 4/16 (or 1/4) because you have HHHT, HHTH, HTHH, THHH, or 4C3Now the probability of 4 heads is 1/16, because the only option is HHHH.Add 6/16 + 4/16 + 1/16 = 11/16 or 68.75%B) The probability of at least two heads is 11/16 or 68.75%.To find the probability of at most two heads add the probability of 1 head and no heads to your answer for exactly two heads. But because heads and tails are equally likely this is symmetrical. So P(1 Head) is the same as P(3 Heads) because 3 Heads means getting exactly one tail and getting one tail should have exactly the same probability as getting exactly one head. Likewise getting 4 heads has the same probability as 0 heads.C) The probability of at most two heads is 11/16 or 68.75%.

A 1-inch-diameter coin is thrown on a table covered with a grid of lines two inches apart. What is the probability the coin lands in a square without touching any of the lines of the grid?

25%Imagine only having to avoid parallel lines instead of a grid.The coin will randomly land either between the lines or across them. The coin has 1 inch of wiggle room between each of the 2 adjacent lines, and 1 inch of wiggle room for the space it occupies when it is touching a line, so with parallel lines (not yet a grid) we have a 50% probability of touching a line.Using the grid now you have the same spacing of 50% which will remove 50% of the remaining area.Now there is a 25% chance the coin will land on a horizontal line, another 25% chance it will land on a vertical line, another 25% chance it will touch both a horizontal and a vertical line, and a 25% chance that it will land between the lines touching neither.

If four coins are tossed simultaneously, what are the possible outcomes?

I like to start with a tree diagram:There are 16 equally likely outcomes:HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTTTHHH THHT THTH THTT TTHH TTHT TTTH TTTT

4 coins are tossed simultaneously. What is the probability of getting 2 heads and 2 tails?

Assuming a fair coin, independent tosses and [math]0[/math] chance of landing on the edge.There are [math]16[/math] possible results. [math]{{4} \choose {2}} = 4*3/2 = 6[/math] of them have 2 heads. Since all 16 are equally likely, the chance is [math]6/16 = 3/8 = 0.375.[/math]

A ball thrown horizontally from the top of a building 55m high strikes the ground at a point 35m from the building. What is the (a) time to reach the ground, (b) the initial speed of the ball, and (c) the velocity by which the ball will strike the ground?

For the first bit you can use [math]s=\frac{1}{2}gt^2[/math]so [math]55=\frac{1}{2}*9.8*t^2[/math][math]t^2=11.22\,\,\,\, so\,\,\,\,\, t=3.35s[/math]The key to these type of questions is finding t first like this because it is the same for the vertical and horizontal components.For part (B) you can use [math]s=ut+\frac{1}{2}at^2[/math]Since acceleration a is zero in the horizontal component this becomes [math]s=ut[/math]We know s (given) and we know t (from part A) so you can go on and calculate u. This is the horizontal component of velocity [math]v_x.[/math]For the last part we consider first the vertical component of motion so we can use [math]v^2=u^2+2as[/math][math]u=0[/math][math]a=9.8\,m.s^{-2}[/math][math]s=55m[/math]You can get [math]v_y[/math]Now we know both horizontal and vertical components we can find the resultant using Pythagoras:[math]v_{res}^{2}=v_{x}^{2}+v_{y}^{2}[/math]So a tip when handling ballistic type questions like these is to find t first as this is the common factor between the horizontal and vertical components of the particle's trajectory, then use the other equations of motion.

Two dice are thrown. What is the probability that the sum of the two is a prime number?

Prime numbers that can be the sum of two fair, 6-sides dice are limited to the range of 2–12.That limits them to sums of 2, 3, 5, 7, and 11. There are there are 36 possible outcomes when you roll two of these dice.out of those 36 possible outcomes, there’s 1 way to make a sum of 2, and 1 way to make a sum of 12.There are 2 ways to make 3, and 2 ways to make 11.3 ways to make 4, and 3 ways to make 10.4 ways to make 5, and 4 ways to make 9.5 ways to make 6, and 5 ways to make 8.Finally there 6 ways to make 7.Add all the ways to make 2, 3, 5, 7, or 11.That would be 1+2+4+6+2= 15 ways to make a prime number sum.15 out of 36 of the outcomes are a prime number sum.So that is the probability ….15/36 = 5/12.

A stone is dropped from the top of a tower.If it hits the ground after 10 seconds, what is the height of the tower?

We can use the formula S=ut+([math]1/2)[/math]a[math]t^2[/math]Where u is initial velocity, ‘s’ is distance travelled (Height of the tower in this case), ‘t’ is time travelled and ‘a’ is the constant acceleration.So , here as the object is dropped , the initial velocity ,u is 0Acceleration is the acceleration due to gravity ‘g’ which is constant and has value of 9.8m/[math]s^2[/math]Time travelled is given as t=10 secondsSubstituting the values in the equation s=ut+([math]1/2)[/math]a[math]t^2 [/math]we gets=0 * 10 + ([math]1/2)[/math]* 9.8 * 10 *10s=490 metersSo, the height of the tower is 490 meters

If we drop a metal ball and a wooden ball from same height at the same time, then which ball will reach the ground first?

The answer to this question will vary based on the situation. Let me elaborate it situation wiseExperiment conducted in total vacuum : In this case both the metal ball and the wooden ball will hit the ground at the same time!!!, irrespective of their shape, weights and sizes. The time required to hit the ground, depends on the acceleration of the object, which in this case if ‘g’ (gravity) acting same on both the objects.Experiment conducted in atmosphere: Here we have air which fills in the atmosphere. Air being a fluid, would enable two things - Buoyancy and air-resistance. In this case, the outcome of the experiment would depend on the following parametersShape of the balls - The ball with a more aerodynamic shape would touch the ground first. As it would have lesser air-resistance, thus the resultant acceleration acting on it would be more.Volume of the ball - The ball having lesser volume will hit the ground first. Since air is a fluid, buoyancy would act on it. Buoyancy is directly proportional to the volume ( ρ x V x g).Weight of the ball - The ball having higher weight will hit the ground first. This is because of the resistive force would have lesser impact on the net acceleration (g-a).In theory, as per Galileo, the answer is indeed straight forward! But life is not that straight forward.I hope this helps!