binomial distribution

hey guys...can u please help me out.
the question is: if there is a probability of 0.2 failure to get through in any attempt to make a telephone call,calculate the most probable number of failures in 10 attempts, also the probability of 3 or more failures in 10 attempts.

abhay said:

hey guys...can u please help me out.
the question is: if there is a probability of 0.2 failure to get through in any attempt to make a telephone call,calculate the most probable number of failures in 10 attempts, also the probability of 3 or more failures in 10 attempts.

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Hint: Is this a binomial distribution?

abhay said:

hey guys...can u please help me out.
the question is: if there is a probability of 0.2 failure to get through in any attempt to make a telephone call,calculate the most probable number of failures in 10 attempts, also the probability of 3 or more failures in 10 attempts.

Click to expand...

Lets look at the first question a bit (actually some like it). Suppose you had a fair coin and flipped it 20 times. In the long run you would expect this to turn up heads [or tails] about 1/2 or 50% of the time, so the expected value would be half of 20. suppose it was a biased coin with probibility of getting a head of 0.4. Then you would expect it to come up heads about 40% of the time in the long run or the expected value for heads is 40% of the number of tosses [60% for tails] or 8 times. Succeding on a telephone call or not is like flipping a coin. Now, can you use this or do you need to prove it?

For part two, you can calculate the probability of 3 failures p³ (1-p)⁷ times the number of combinations the three failures could be arranged, ₁₀C₃. The do the same for 4, then 5, then 6, ..., then 10 and add them all up. Or be lazy, like me, and just 0, 1, and 2, add them up and that will given the probability that there are less than three. And what is the probability that there is 3 or more but the opposite of there are less than 3.