Probability_Problem_Solution_Series1.ipynb

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    "### First some fundamental formulas\n",
    "\n",
    "**The combination of n different things taken r at a time is denoted by _nCr_ and used the formula for calculation is below**\n",
    "\n",
    "![](https://imgur.com/sZmjKsC.png)\n",
    "\n",
    "\n",
    "**If two dice are rolled, what is the probability that the sum of the upturned faces will equal 7?**\n",
    "\n",
    "**Solution** — We shall solve this problem under the assumption that all of the 36 possible outcomes are equally likely. Since there are 6 possible outcomes-namely, (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1)-that result in the sum of the dice being equal to 7, the desired probability is 6/36 = 1/6\n",
    "\n",
    "**_Problem -1 asked at a Google Interview:_**\n",
    "\n",
    "> **If the probability of seeing a car on the highway in 30 minutes is 0.95, what is the probability of seeing a car on the highway in 10 minutes? (assume a constant default probability)**\n",
    "\n",
    "Some clarifications that can help are:\n",
    "\n",
    "“Seeing a car” means “seeing at least 1 car.”\n",
    "\n",
    "“Constant default probability” basically means there is a constant probability rate of seeing a car (there’s no picking of probability during rush hour traffic, for example). Mathematically this can be stated as —Probability of observing a car in any given non-overlapping time interval of equal length are equal and independent.\n",
    "\n",
    "With this assumption, I can think of the probability of seeing a car during a time interval is like a coin flip event, which also occurs with “Constant default probability”\n",
    "\n",
    "So, now, think about each 10-minute interval like a coin flip. Suppose the probability of not seeing a car in 10 minutes is p.\n",
    "\n",
    "**Now first recount the Probability Multiplication Rules**\n",
    "\n",
    "The multiplication rule is about the probability of two events happening at the same time. There are two multiplication rules. The general multiplication rule formula is:\n",
    "\n",
    "#### P(A ∩ B) = P(A) * P(B|A)\n",
    "\n",
    "means “the probability of A happening given that B has occurred”.\n",
    "\n",
    "If the two events are independent. In other words, it only works if one event does not change the probability of the other event then,\n",
    "\n",
    "### P(A and B) = P(A) * P(B)\n",
    "\n",
    "**Let’s divide 30 minutes time intervals into three 10 minutes intervals as A, B, and C.**\n",
    "\n",
    "P(not A) = Probability of not seeing a car in the first 10 minutes\n",
    "\n",
    "P(not B) = Probability of not seeing a car in the second 10 minutes\n",
    "\n",
    "P(not C) = Probability of not seeing a car in the third 10 minutes\n",
    "\n",
    "As all are independent events with constant default Probability, so\n",
    "\n",
    "P(not A) = P(not B) = P(not C) = lets say **p**\n",
    "\n",
    "**Then what is the probability of not seeing a car in 20 minutes? This is the probability of not seeing a car in 10 minutes, followed by the same event.**\n",
    "\n",
    "**Pr(no cars in 20) = Pr(no cars in 10) Pr(no cars in 10)**\n",
    "\n",
    "**= p * p = p²**\n",
    "\n",
    "Similarly, the probability of not seeing a car in 30 minutes is the probability of not seeing a car in 20 minutes, followed by the event of not seeing a car in another 10 minutes.\n",
    "\n",
    "**Pr(no cars in 30) = Pr(no cars in 20) Pr(no cars in 10) = p² * p = p³**\n",
    "\n",
    "Now the probability of seeing at least 1 car in 30 minutes can be calculated as the complement event, equal to 1 minus the probability of seeing no cars in 30 minutes.\n",
    "\n",
    "Pr(at least 1 car in 30) = 1 — Pr(no cars in 30)\n",
    "Pr(at least 1 car in 30) = 1 — p³\n",
    "\n",
    "We are given this is equal to 0.95. We can solve for p:\n",
    "\n",
    "1 — p³ = 0.95\n",
    "p³ = 0.05\n",
    "p = (0.05)^1/3 ≈ 0.368\n",
    "\n",
    "This is the probability of seeing no cars in 10 minutes. The probability of seeing at least 1 car is the complement event, equal to 1 minus the probability of seeing no cars in 10 minutes.\n",
    "\n",
    "Pr(at least 1car in 10) = 1 — Pr(no cars in 10)\n",
    "Pr(at least 1 car in 10) = 1 — p\n",
    "Pr(at least 1 car in 10) ≈ 0.632\n",
    "\n",
    "**So the final answer** — the probability of seeing at least 1 car on the highway in 10 minutes is approximately 63 percent.\n",
    "\n",
    "**_Probability Problem-2_**\n",
    "\n",
    "First the formulae of The Multinomial Distribution\n",
    "\n",
    "![](https://imgur.com/KHKd8W5.png)\n",
    "\n",
    "![](https://imgur.com/Ci0cgYx.png)\n",
    "\n",
    "![Imgur](https://imgur.com/Xpp6QK8.png)\n",
    "\n",
    "\n",
    "This distribution, which is a generalization of the binomial distribution, is called the multinomial distribution since the above equation is the general term in the multinomial expansion of\n",
    "\n",
    "![Imgur](https://imgur.com/ilJrRvE.png)\n",
    "\n",
    "\n",
    "---\n",
    "\n",
    "**_Probability Problem-3_**\n",
    "\n",
    "A box contains 5 red balls, 4 white balls, and 3 blue balls. A ball is selected at random from the box, its color is noted, and then the ball is replaced. Find the probability that out of 6 balls selected in this manner, 3 are red, 2 are white, and 1 is blue\n",
    "\n",
    "**Method 1 (by formula):**\n",
    "\n",
    "Lets first calculate the total Probabilities of choosing a colored ball out of Total 12\n",
    "\n",
    "P(Total prob of red at any drawing) => 5 / 12\n",
    "\n",
    "P(Total prob of white at any drawing) => 4/12\n",
    "\n",
    "P(Total prob of blue at any drawing) => 3/12\n",
    "\n",
    "Now the formulae we are going to apply is that of **Finding the Number of Permutations of n Non-Distinct Objects** which is\n",
    "\n",
    "![](https://imgur.com/HpePJkw.png)\n",
    "\n",
    "\n",
    "And also the by the **Multiplication theorem of probability / Formula for Joint Probability**\n",
    "\n",
    "If A and B are independent events, we can reduce the formula to\n",
    "\n",
    "**_P(A and B) = P(A) * P(B)_**\n",
    "\n",
    "This means the probability of choosing any red ball is 5 / 12. Then the probability of choosing 3 red balls is (5 /12)^3\n",
    "\n",
    "So, then the required probability for our problem is\n",
    "\n",
    "![](https://imgur.com/94IrIcu.png)\n",
    "\n",
    "\n",
    "---\n",
    "\n",
    "**Explanations**\n",
    "\n",
    "The probability of choosing any red ball is 5 /12. Then the probability of choosing 3 red balls is (5 /12)^3.\n",
    "Similarly, the probability of choosing 2 white balls is (4/ 12)^ 2, and of choosing 1 blue ball, (3 12)^1.\n",
    "\n",
    "And therefore, the probability of choosing 3 red, 2 white, and 1 blue in that order is\n",
    "\n",
    "![](https://imgur.com/kdc32D3.png)\n",
    "\n",
    "\n",
    "But the same selection can be achieved in various other orders, and the number of these different ways is\n",
    "\n",
    "Then the required probability is\n",
    "\n",
    "![](https://imgur.com/LXusO1A.png)\n",
    "\n",
    "\n",
    "**Method 2 (by the law of multinomial expansion**) **:**\n",
    "\n",
    "The required probability is the term\n",
    "\n",
    "![](https://imgur.com/ceJgqYT.png)\n",
    "\n",
    "\n",
    "in the multinomial expansion of\n",
    "\n",
    "![](https://imgur.com/XWh7R9Q.png)\n",
    "\n",
    "\n",
    "where\n",
    "\n",
    "p_r = 5 /12,\n",
    "\n",
    "p_w =4/ 12, and\n",
    "\n",
    "p_b =3/12.\n",
    "\n",
    "By actual expansion, the above result is obtained."
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